C:/Documents and Settings/SAVA DASHEV/My … manual3.pdf · 3.7.2 BBb3: Simplify negative powers of sums and differences. . . . . . . . 66 3.7.3 BBb4: ... 2 (x+1). 4. CHAPTER 3. ALGEBRA

Contents

3 Algebra 3

3.1 Rational expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3.1.1 BAa1: Reduce to lowest terms . . . . . . . . . . . . . . . . . . . . . . 3

3.1.2 BAa2: Add, subtract, multiply, and divide . . . . . . . . . . . . . . . 5

3.1.3 BAa3: Find and simplify products of the form(ax + b)(c/x + d). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.1.4 BAa4: Evaluate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3.1.5 BAa5: Simplify complex fractions . . . . . . . . . . . . . . . . . . . . 9

3.2 Factoring quadratics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3.2.1 BAb1: Factor a quadratic trinomial in two variables . . . . . . . . . . 11

3.2.2 BAb3: Factor a perfect square trinomial in more than one variable . 13

3.2.3 BAb4: Recognize relationships among coefficients of a factorabletrinomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.3 Operations on Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.3.1 BAe1: Simplify polynomial expressions involving products of bino-mials in one or more variables . . . . . . . . . . . . . . . . . . . . . . 18

3.3.2 BAe2: Factor a difference of squares . . . . . . . . . . . . . . . . . . . 19

3.3.3 BAe4: Divide a third degree by a first or second degree to obtain thequotient and remainder . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.3.4 BAe5: Raise a binomial to the second or third power . . . . . . . . . 22

3.4 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.4.1 BAc2: Find the slope of a line from an equation . . . . . . . . . . . . 24

3.4.2 BAc3: Find the slope of a line from the coordinates of two points. . . 26

3.4.3 BAc4: Find the equation of a line from the coordinates of two points. 28

3.4.4 BAc5: Find the equation of a line given 1 point and the slope. . . . . 31

3.4.5 BAc6: Determine and/or plot the intercepts of a line, given an equa-tion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.4.6 BAc9: Given an equation determine if a point is above or below thegraph. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.5 Quadratic/Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.5.1 BAd1: Quadratics of the form ax2 + b = 0 or (ax + b)2 + c = 0. . . . . 38

3.5.2 BAd2: Quadratics of the form ax2+bx+c = 0 with a 6= 1 by factoringover the integers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

1

3.5.3 BAd3: Quadratics using the quadratic formula, where the discrimi-nant might be positive, negative or zero. . . . . . . . . . . . . . . . . 43

3.5.4 BAd4: Equations quadratic in u, where u = x2, x3, .... . . . . . . . . . 463.5.5 BAd5: Rational equations leading to quadratic equations. . . . . . . 493.5.6 BAd6: Rational equations leading to linear equations. . . . . . . . . . 523.5.7 BAd7: Solve literal equations involving rational expressions. . . . . . 54

3.6 Radicals and Fractionial exponents . . . . . . . . . . . . . . . . . . . . . . . . 563.6.1 BBa4: Express radicals in terms of fractional exponents and vice versa. 563.6.2 BBa1: Evaluate numerical expressions involving radicals and/or

fractional exponents. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.6.3 BBa2: Simplify radical expressions. . . . . . . . . . . . . . . . . . . . . 583.6.4 BBa3: Rationalize the numerator or denominator of a quotient in-

volving radicals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.6.5 BBa5: Multiply radical expressions (monomials, binomials). . . . . . 613.6.6 BBa6: Simplify and combine like radicals. . . . . . . . . . . . . . . . . 623.6.7 BBa7: Simplify expressions involving radicals by changing to frac-

tional exponents and using exponent laws. . . . . . . . . . . . . . . . 633.7 Negative Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.7.1 BBb2: Use reciprocal rule to eliminate denominator negative expo-nents. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.7.2 BBb3: Simplify negative powers of sums and differences. . . . . . . . 663.7.3 BBb4: Evaluate or simplify expressions involving negative exponents. 67

3.8 Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693.8.1 BBc1: Evaluate numerical expressions involving absolute values. . . 693.8.2 BBc2: Recognize the graph of y = |x| and its variations. . . . . . . . . 703.8.3 BBc3: Solve equations of the type |ax + b| + c = d. . . . . . . . . . . . 74

3.9 Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 763.9.1 BBd1: Solve or graph linear inequalities in one variable. . . . . . . . 763.9.2 BBe1: Solve consistent 2 x 2 systems by substitution or elimination. . 833.9.3 BBe2: Interpret 2 x 2 systems graphically . . . . . . . . . . . . . . . . . 873.9.4 BBe3: Recognize inconsistent or dependent systems algebraically or

graphically. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 913.10 Verbal Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

3.10.1 BBf1: Work/rate. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 933.10.2 BBf2: Distance/rate/time. . . . . . . . . . . . . . . . . . . . . . . . . . 973.10.3 BBf3: Mixture. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1003.10.4 BBf4: Number (reciprocal, digit, consecutive, integer, etc.). . . . . . . 1033.10.5 BBf5: Area and perimeter. . . . . . . . . . . . . . . . . . . . . . . . . . 1073.10.6 BBf6: Other miscellaneous problems. . . . . . . . . . . . . . . . . . . 1103.10.7 BBf5: Area and perimeter. . . . . . . . . . . . . . . . . . . . . . . . . . 113

2

Chapter 3

Algebra

3.1 Rational expressions

3.1.1 BAa1: Reduce to lowest terms

We need to recall the rule for simplifying fractions. To do so, we need to factor out com-mon factors and then we can divide out common factors. Doing it any other way is notsupported by the rules of arithmetic.

1.x(x + 2)

x(2x + 4)The numerator is factored. We need to factor out the denominator. We see that thesecond factor contains common factor 2, so we factor it out and divide all commonfactors x(x + 2), assuming x(x + 2) 6= 0:

=x(x + 2)

2x(x + 2)

=x(x + 2)

x(x + 2)× 1

2=

1

2.

2.x2(x2 + 2)

x(2x2 + 4)We factor out 2 from the denominator, then we divide out all common factors on thesecond step. We can do it only if x 6= 0

=x2(x2 + 2)

2x(x2 + 2)

=x(x2 + 2)

x(x2 + 2)× 1

2x

=1

2x.

3

3.1. RATIONAL EXPRESSIONS CHAPTER 3. ALGEBRA

3.(x + 9)(x − 2)

x(3x + 27)We factor 3 from the denominator. Then we divide out all common factors, shownon separate fractions and divide out them (assuming x − 9 6= 0).

=(x + 9)(x − 2)

x[3(x + 9)]

=x + 9

x + 9× x − 2

3x

=x − 2

3x.

4.a(3x − 2)

(x + 3)(6 − 9x)We observe if we reverse the signs in the parenthesis and factor out 3, we may havecommon factors to divide out. We do so and have:

=a(3x − 2)

(x + 3)(−3)(3x − 2)

We divide out (3x − 2) assuming that (3x − 2) 6= 0.

= − a

3(x + 3).

5.x(x2 + 2x + 1)

x(2x + 2)We can factor the numerator further seeing that x2 + 2x + 1 = (x + 1)2. There is onlyone factor we can factor out from the denominator - the number 2. On the secondstep, we divide out all common factors.

=x(x + 1)2

x2(x + 1)

=x(x + 1)

x(x + 1)× 1

2(x + 1)

We assume x(x + 1) 6= 0.

=1

2(x + 1).

4

CHAPTER 3. ALGEBRA 3.1. RATIONAL EXPRESSIONS

3.1.2 BAa2: Add, subtract, multiply, and divide

To add or subtract rational expression, we need to find the Least Common Denominator.We multiply all expressions by the appropriate factors and then we add or subtract theresulting numerators.

1.(x + 2)

x(x + 4)+

x

2The LCD is 2x(x + 4).

=

(

(x + 2)

x(x + 4)× 2

2

)

+

(

x

2× x(x + 4)

x(x + 4)

)

=2(x + 2) + x[x(x + 4)]

2x(x + 4)

=x3 + 4x2 + 2x + 4

2x(x + 4).

2.(x + 3)

x(x + 4)− x

2=

LCD = 2x(x + 4)

=(x + 3)2

2x(x + 4)− x[x(x + 4)]

2x(x + 4)

=2x + 6 − x3 − 4x2

2x(x + 4)

=−x3 − 4x2 + 2x + 6

2x(x + 4).

3.(x + 3)

x(x + 4)× x

2 + x

=(x + 3)x

x(x + 4)(x + 2)

We divide out a common factor x, assuming (x 6= 0).

=x + 3

(x + 4)(x + 2).

4.(x + 3)

x(x + 4)÷ x

2 + x=

=(x + 3)

x(x + 4)× (2 + x)

x=

=(x + 3)(x + 2)

x2(x + 4).

5


5.(x − 3)

x(x2 + 4)− x

2 + xLCD = x(x2 + 4)(2 + x)

=(x − 3)(2 + x)

x(x2 + 4)(2 + x)− x([x(x2 + 4)]

x(x2 + 4)(2 + x)

=x2 − x − 6 − x4 − 4x2

x(x + 2)(x2 + 4)

=−x4 − 3x2 − x − 6

x(x + 2)(x2 + 4)

= −x4 + 3x2 + x + 6

x(x + 2)(x2 + 4).

Note we put the minus sign in front of the fraction and all minuses were replacedby pluses.

6


3.1.3 BAa3: Find and simplify products of the form

(ax + b)(c/x + d).

We need to multiply expressions containing two terms. A convenient way of remember-ing how to do it is using FOIL method. We will show it in detail for # 1.

1. (3x + 2)

(

2

x+ 5

)

=

(

3x × 2

x

)

+ (3x × 5) +

(

2 × 2

x

)

+ (2 × 5)

= 6 + 15x +4

x+ 10

= 15x + 16 +4

x.

2. (x + 2)

(

1

x− 2

)

= 1 − 2x +2

x− 4

= −2x − 3 +2

x.

3. (5x − 7)

(

4

3x+ 5

)

=20

3+ 25x − 28

3x− 35

= 25x − 85

3− 28

3x.

4. (−x + 2)

(

7

x− 5

)

= −7 + 5x +14

x− 10

= 5x − 17 +14

x.

5. (7x − 2)

(−2

x+ 15

)

= −14 + 105x +4

x− 30

= 105x − 44 +4

x.

7


3.1.4 BAa4: Evaluate

1. If x = 2, then

(

(2 − x)

x − 5

)

=2 − (+2)

(+2) − 5

=0

−3= 0.

2. If x = −2, then

(

(2 − x)

x − 5

)

=2 − (−2)

(−2) − 5

=4

−7= −4

7.

3. If x = 5, then

(

(2 − x)

x2 − 5

)

=2 − (+5)

(+5)2 − 5

=−3

20= − 3

20.

4. If x = −7, then

(

(2x2 − x)

x − 5

)

=2(−7)2 − (−7)

(−7) − 5=

105

−12= −35

4.

5. If x = −3, then

(

(2 − x2)

x2 − 5x

)

=2 − (−3)2

(−3)2 − 5(−3)

=2 − 9

9 + 15= − 7

24.

Note the use of parentheses for negative numbers. This way you will not forget to changethe sign if you need to.

8


3.1.5 BAa5: Simplify complex fractions

We can look at the numerator of the complex fraction as a number, which we try to di-vide by the denominator fraction. Dividing by a fraction means that we multiply by thereciprocal of that fraction. This step is shown only in # 1.

1. Reduce:(

(x+2)(x+5)x+3

)

(

(x+2)(x+5)x+4

)

=(x + 2)(x + 5)

x + 3÷ (x + 2)(x + 5)

x + 4

=(x + 2)(x + 5)

x + 3× x + 4

(x + 2)(x + 5)

=x + 4

x + 3.

2. Reduce:(

(x2+2)(x+5)x+3

)

(

x+2(x+4)(x+1)

)

=(x2 + 2)(x + 5)

x + 3÷ x + 2

(x + 4)(x + 1)

=(x2 + 2)(x + 5)

x + 3× (x + 4)(x + 1)

x + 2

=(x2 + 2)(x + 5)(x + 4)(x + 1)

(x + 3)(x + 2).

3. Reduce:(

x(x+2)(x2+5)x+3

)

(

x+2x(x+4)

)

=x(x + 2)(x2 + 5)

x + 3÷ x + 2

x(x + 4)

We factor out the numerator of the numerator fraction.

=x(x + 2)(x2 + 5)

x + 3× x(x + 4)

x + 2

=x2(x2 + 5)(x + 4)

x + 3.

9


4. Reduce:(

(x2+2x)(x+5)x+3

)

(

x+2x2+4x

)

=(x2 + 2x)(x + 5)

x + 3÷ x + 2

x2 + 4x

=x(x + 2)(x + 5)

x + 3× x(x + 4)

x + 2

=x2(x + 5)(x + 4)

x + 3.

5. Reduce:(

(x−2)(x+5)x+3

)

(

−x+2x+4

)

=(x − 2)(x + 5)

x + 3÷ −x + 2

x + 4

=(x − 2)(x + 5)

x + 3× x + 4

−x + 2

=−(−x + 2)(x + 5)(x + 4)

(x + 3)(−x + 2)

= −(x + 5)(x + 4)

x + 3.

10

CHAPTER 3. ALGEBRA 3.2. FACTORING QUADRATICS

3.2 Factoring quadratics

3.2.1 BAb1: Factor a quadratic trinomial in two variables

To factor quadratics, one way is to use FOIL method. We look for factors of the format(A1a + B1b)(A2a + B2b). Evaluating this, we obtain:

(A1a + B1b)(A2a + B2b) = A1A2a2 + (A1B2 + A2B1)ab + B1B2b

2

We organize the information in a table. We will show the process in #1.

1. a2 + 5ab + 6b2

Looking at the above expression we can see A1A2 = 1. One natural guess is A1 =A2 = 1.We replace A1 and A2 in the sum A1B2 + A2B1 = 1B2 + 1B1 = B2 + B1. The sumshould be equal to 5.We finally look at the product B1B2. It must be equal to 6. On the first step, we factorproduct B1B2 = 6. The sum is (+5), so we need to look only at positive factors. Thepossibilities for the factors are (+6, +1) and (+3, +2). The first pair of factors hassum (+7) and second one has sum (+5). We see the factors must be (+3) and (+2).All the information is organized in the table below. In the left column are shown thetwo factors of the product in the top row. Their sum is shown in the right column.

Product=6 Sum of factors+1, +6 7 NO!+2, +3 5 YES!

= a2 + 5ab + 6b2 = 1 × 1 × a2 + (1 × 2 + 1 × 3)ab + 2 × 3 × b2

= a2 + 2ab + 3ab + 6b2

= a(a + 2b) + 3b(a + 2b)

= (a + 2b)(a + 3b).

2. a2 − 5ab + 6b2

The only difference between # 2 and # 1 is that the sum B1 + B2 = −5, i.e. negativefive. It means that the factors are negative. The table shows the values:

Product=6 Sum of factors−2, −3 −5 YES!

= a2 − 2ab − 3ab + 6b2

We combine first and second terms, and third and forth terms; then factor:

= a(a − 2b) − 3b(a − 2b)

= (a − 2b)(a − 3b).

11

3.2. FACTORING QUADRATICS CHAPTER 3. ALGEBRA

3. a2 + 4ab − 5b2 =

Product=−5 Sum of factors+1, −5 −5 NO!−1, +5 +5 YES!

= a2 + 5ab − ab − 5b2

= a(a + 5b) − b(a + 5b)

= (a + 5b)(a − b).

4. 6a2 + 5ab + b2

We rearrange this in the following fashion

= b2 + 5ab + 6a2

We can now to use the table from #1

Product=6 Sum of factors+1, +6 7 NO!+2, +3 5 YES!

= b2 + 2ab + 3ab + 6a2

= b(b + 2a) + 3a(b + 3a)

= (b + 2a)(b + 3a).

5. a2 + 12ab + 20b2

Product=20 Sum of factors1, 20 21 NO!2, 10 12 YES!

= a2 + 2ab + 10ab + 20b2

= a(a + 2b) + 10b(a + 2b)

= (a + 2b)(a + 10b).

12


3.2.2 BAb3: Factor a perfect square trinomial in more than one variable

We need to recognize the perfect square formula:

A2 + 2AB + B2 = (A + B)2

and

A2 − 2AB + B2 = (A − B)2 = (B − A)2

The last identity resembles the property that the opposite numbers have one and the samesquare.Example: (−3)2 = 9 = 32

1. a2 − 6ab + 9b2

We see there are differences of numbers squared:A2 = a2,B2 = 9b2 = (3b)2,−2AB = −6ab = −2 × a(3b).Then:

= a2 − 2a(3b) + (3b)2

= (a − 3b)2.

We can check the proper factoring by remultiplying:

(a − 3b)2 = a2 − 2a × (3b) + (−3b)2 = a2 − 6ab + 9b2.

2. 4a2 − 12ab + 9b2

A2 = (2a)2,B2 = 9b2 = (3b)2,−2AB = −12ab = −2(2a)(3b).Then:

= (2a)2 − 2(2a)(3b) + (3b)2

= (2a − 3b)2.

3. 25a2 + 20ab + 4b2 =

A2 = (5a)2,B2 = 4b2 = (2b)2,2AB = 20ab = 2(5a)(2b).Then:

= (5a)2 + 2(5a)(2b) + (2b)2

= (5a + 2b)2.

13


4. a2 − 2ab−1 + b−2

A2 = a2,B2 = b−2 = (b−1)2,−2AB = −2ab−1.

Then:= a2 − 2ab−1 + b−2

= (a − b−1)2.

This is a very important example. We see the square here is b−2 = (b−1)2 and a2. Theequivalent way you could see this is a2 − 2 × a/b + 1/b2. We can say:

a2 − 2ab−1 + b−2 = a2 − 2a

b+

1

b2.

Then:

(a − b−1)2 = (a − 1

b)2.

5. 36a2 + 36ab + 9b2

A2 = 36a2 = (6a)2,B2 = 9b2 = (3b)2,2AB = 36ab = 2(6a)(3b).Then:

= (6a)2 + 2(6a)(3b) + (3b)2

= (6a + 3b)2.

Note! Parentheses are used in all the expressions above. This is first, so we do not getconfused, and second, to show not only the variable but also the whole expression inparenthesis is squared. In the cross product, it shows the separated factors to you.

In all cases, if you feel you could be confused by notation, use parenthesis for indicat-ing what arithmetic operation you need to do first.

14


3.2.3 BAb4: Recognize relationships among coefficients of a factorable

trinomial

Find a value of b so that the quadratic factors with integer coefficients, and give the fac-torization. There may be more than one correct answer. You only need give one correctanswer.I will give you all possible answers. For #2 and #4, there are infinitely many of them. Theway these could be computed will be shown in a note at the end of the section for theinterested reader. The other items have a finite number of solutions.

1. x2 + (5 + b)x + 12.We can use FOIL to find all the answers it this case. We look for whole numberswhich have product 12 and sum (5+ b). So, we factor 12 and show all groups of twofactors we may have: (1, 12), (2, 6), (3, 4); we could have negative factoring as well:(−1,−12), (−2,−6), (−3,−4). For each pair of factors, we can compute the sum ofthe factors, it must be (5 + b). We then will have an equation we can solve for b:

5 + b = 1 + 12,

b = 8.

In a table, we show all possible values for b. If your answer is not among them,check your work.

Product=12 Sum=5 + b b1, 12 13 82, 6 8 33, 4 7 2-1, -12 -13 -18-2, -6 -8 -13-3, -4 -7 -12

2. x2 + 7x + 2b.

We use FOIL method again. Let us have factors

x2 + 7x + 2b = (x + u)(x + v) = x2 + vx + ux + uv = x2 + (u + v)x + uv.

We choose one of the factors to be u = 3. Then, using the fact that the left and theright side are simply two representations of the same trimoial, so u + v = 7 anduv = 2b. We find v first.

v = 7 − u = 7 − 3 = 4.

Second, we find b.2b = uv = 3 × 4 = 12,

15


b = 6.

One solution is x2 + 7x + 2b = (x + 3)(x + 4) if b = 6.

3. x2 + (7 − b)x + 24.

We use the same (FOIL) method as in #1. The table below shows the values of b forwhich the trinomial is factored.

Product=24 Sum=7 − b b1, 24 25 -182, 12 14 -73, 8 11 -44, 6 10 -3-1, -24 -25 32-2, -12 -14 21-3, -8 -11 18-4, -6 -10 17

Example for the first row we have b = (−18). Then:

x2 + [7 − (−18)]x + 24 = x2 + 25x + 24 = (x + 1)(x + 24).

4. x2 + x + 9b.

We applay the same method as in #2:

x2 + x + 9b = (x + u)(x + v) = x2 + vx + ux + uv = x2 + (u + v)x + uv.

We can choose u = 2. Then, v = −1. We find then that uv = 9b, or

b =uv

9=

2 × (−1)

9= −2

9.

There is no restriction on b to be integer. We check:

x2 + x + 9 × −2

9= x2 + x − 2 = (x + 2)(x − 1).

Note, that even b is not an integer, the quadratic trinomial still have integer coeffi-cients.

5. x2 + (1 − b)x + 15In the table is shown the solution using the same idea.

16


Product=15 Sum=1 − b b1, 15 16 -153, 5 8 -7-1, -15 -16 17-3, -5 -8 9

Check (using the last row):

x2 + (1 − 9)x + 15 = x2 − 8x + 15 = (x − 3)(x − 5).

Finding the solutions for #2: we can choose u ∈ N , then from u + v = 7 we find

v = 7 − u and using FOIL, we compute:

u(7 − u) = 2b,

b =u(7 − u)

2.

We have:

x2 + 7x + 2 × u(7 − u)

2

= x2 + [u + (7 − u)]x + u(7 − u) = (x + u)[x + (7 − u)].

17

3.3. OPERATIONS ON POLYNOMIALS CHAPTER 3. ALGEBRA

3.3 Operations on Polynomials

3.3.1 BAe1: Simplify polynomial expressions involving products of bi-

nomials in one or more variables

1. (x − 1)(x + 2) + x(x − 3)We multiply the two binomials using FOIL and expand the second product. On thenext step, we combine all the like terms.

= x2 + 2x − x − 2 + x2 − 3x

= (1 + 1)x2 + (2 − 1 − 3)x − 2

= 2x2 − 2x − 2.

2. (x − 1)(x + 2) + x(x − 3)

= x2 + 2x − x − 2 + x2 − 3x

= (1 + 1)x2 + (2 − 1 − 3)x − 2

= 2x2 − 2x − 2.

3. (x − 1)(x2 + 2) + x(x2 − 3)This is again FOIL method. The only difference is one of the binomials is quadratic.We will not have like terms expanding this product. However, when expandingsecond product, we may find like terms:

= x3 + 2x − x2 − 2 + x3 − 3x

= (1 + 1)x3 − x2 + (2 − 3)x − 2

= 2x3 − x2 − x − 2.

4. (x − 2)(x + 2) − x(x − 3)You may notice we have a product of sum and difference. There is a formula whichyou’d like to remember: (A + B)(A − B) = A2 − B2. Here, A = x; B = 2. Check outthe FOIL method used below and see that the crossproducts subtract to 0.

= x2 + 2x − 2x − 4 − x2 + 3x

= 3x − 4.

5. (2x − 1)(x + 7) − x(x − 3)

= 2x2 + 14x − x − 7 − x2 + 3x

= (2 − 1)x2 + (14 − 1 + 3)x − 7

= x2 + 16x − 7.

18

CHAPTER 3. ALGEBRA 3.3. OPERATIONS ON POLYNOMIALS

3.3.2 BAe2: Factor a difference of squares

We will use the formula A2−B2 = (A−B)(A+B). I will give you the factors in each case.Note, that in some cases, we can factor out a common factor which is not a binomial.

1. 25x2 − 81y2

A2 = 25x2 = (5x)2,B2 = 81y2 = (9y)2.We use parenthesis to change the order of arithmetic operations, first we do multiplica-tion , and, then raise to a power.

= (5x)2 − (9y)2

= (5x − 9y)(5x + 9y).

2. 4x2 − 9y2

A2 = 4x2 = (2x)2,B2 = 9y2 = (3y)2.

= (2x)2 − (3y)2

= (2x − 3y)(2x + 3y).

3. 16x2 − 144y2

A2 = 16x2 = 42(x)2,B2 = 144y2 = 16(3y)2.

= 16x2 − 16(3y)2

= 16[x2 − (3y)2]

= 16(x − 3y)(x + 3y).

4. 25x2 − 36y2

A2 = 25x2 = (5x)2,B2 = 36y2 = (6y)2.

= (5x)2 − (6y)2

= (5x − 6y)(5x + 6y).

5. y4x2 − 16a2b6

A2 = y4x2 = (y2x)2,B2 = 16a2b6 = (4ab3)2.In this example we have higher powers than two. We see the square could be apower of a variable. Here, A = y2x, and B = 4ab3.

= (y2x)2 − (4ab3)2

= (y2x − 4ab3)(y2x − 4ab3).

19


3.3.3 BAe4: Divide a third degree by a first or second degree to obtain

the quotient and remainder

The division of polynomials is very similar to long division of numbers.Note when we have missing terms, in either dividend and the divisor, we have to showthese one way or another. First, we can write them with coefficient zero or the other wayis to leave an empty space in the dividend when writing it.

The division in detail (including the way one can check his answer) is shown #1 and#2

1. (x3 − 3x + 4) ÷ (x2 + 2x)The division:

x − 2x2 + 2x)x3 + 0x2 − 3x + 4

−(x3 + 2x2)

−2x2 − 3x−(−2x2 − 4x)

x + 4 reminder

= x − 2 +x + 4

z2 + 2x.

We check the division by multiplication. We multiply the answer and the dividendand then we add the reminder. If we worked correctly, we will obtain the divisor.

answer × divisor + reminder = dividend

(x − 2)(x2 + 2x) + x + 4 = x3 + 2x2 − 2x2 − 4x + x + 4

= x3 − 3x + 4. TRUE!

2. (3x3 + 3x2 + 4) ÷ (x2 − x + 2)The division:

3x + 6x2 − x + 2)3x3 + 3x2 + 0x + 4

−(3x3 − 3x2 + 6x)

6x2 − 6x + 4−(6x2 − 6x + 12)

−8 reminder

= 3x + 6 +−8

x2 − x + 2.

20


Check:answer × divisor + reminder = dividend

(3x + 6)(x2 − x + 2) − 8

= 3x3 − 3x2 + 6x + 6x2 − 6x + 12 − 8

= 3x3 + 3x2 + 4. TRUE!

3. (x3 + 3x + 4) ÷ (x + 2) =

= x2 − 2x + 7 +−10

x + 2.

4. (x3 + 4x2 − 3x + 4) ÷ (x − 3) =

= x2 + 7x + 18 +58

x − 3.

5. (2x3 − 3x2 + 5x − 4) ÷ (x2 − 2) =

= 2x − 3 +9x − 10

x2 − 2.

Once more, here is the way to check the answer. We multiply the quotient and the divisorand add a reminder (example #5):

(x2 − 2)(2x − 3) + 9x − 10 =

= 2x3 − 3x2 − 4x + 6 + 9x − 10 =

= 2x3 − 3x2 + 5x − 4.

We got the original expression ( the dividend), so that we worked properly.

21


3.3.4 BAe5: Raise a binomial to the second or third power

We can use the following formulas for raising to second and third powers:

(A + B)2 = A2 + 2AB + B2.

Using the formula for the sum, and the fact that −B = +(−B) we find

(A − B)2 = [A + (−B)]2 = A2 + 2A(−B) + (−B)2 = A2 − 2AB + B2.

The same way we compute the third power of the sum:

(A + B)3 = A3 + 3A2B + 3AB2 + B3.

Changing B to −B we obtain

(A − B)3 = [A + (−B)]3.

= A3 + 3A2(−B) + 3A(−B2) + (−B)3 == A3 − 3A2B + 3AB2 − B3.

Note that squares are shown with + plus sign regardless whether they are with plusor minus sign in the sum. The only difference when we square a difference is that thecrossproduct has negative sign.Note, that in general we do not need two of the formulas: difference squared and differ-ence cubed because these can be derived from the sum squared and the sum cubed byappropriate change of signs.

1. (x + 1)2

A = x;B = 1.

= x2 + (2 × x × 1) + 1 = x2 + 2x + 1.

(x + 1)3

= x3 + 3 × x2 × 1 + 3 × x × 12 + 13 = x3 + 3x2 + 3x + 1.

2. (x + 2)2

A = x;B = 2.

= x2 + 2 × x × 2 + 22 = x2 + 4x + 4.

(x + 2)3

= x3 + 3 × x2 × 2 + 3 × x × 22 + 23 = x3 + 6x2 + 12x + 8.

22


3. (2x − 1)2

A = 2x;B = −1.

= (2x)2 + 2 × (2x × (−1) + (−1)2 = 4x2 − 4x + 1.

(2x − 1)3

= (2x)3 + 3 × (2x)2 × (−1)1 + 3 × (2x) × (−1)2 + (−1)3 = 8x3 − 12x2 + 6x − 1.

4. (a + 2b)2

A = a;B = 2b.

= a2 + 2a(2b) + (2b)2 = a2 + 4ab + 4b2.

(a + 2b)3

= a3 + 3a2(2b) + 3a(2b)2 + (2b)3 = a3 + 6a2b + 12ab2 + 8b3.

5. (x − 3)2

A = x;B = 3.

= x2 + 2 × x × (−3) + (−3)2 = x2 − 6x + 9.

(x − 3)3

= x3 + 3 × x2 × (−3) + 3 × x × (−3)2 + (−3)3 = x3 − 9x2 + 27x − 27.

23

3.4. GRAPHS CHAPTER 3. ALGEBRA

3.4 Graphs

3.4.1 BAc2: Find the slope of a line from an equation

The slope of a line is a fraction

slope of a line =coeff x

coeff y.

Example: for the equation 3y + 4x = 12 we subtract 4x from both sides.

3y + 4x − 4x = 12 − 4x,

3y = −4x + 12.

The slope is

slope of a line =coeff x

coeff y=

−4

3.

We can divide by 3:

y =−4

3x +

12

3=

−4

3+ 4.

The slope is

slope = (−4

3)/1.

1. The slope of the line y = 5x + 7 is 5.

y =5

1x + 7.

2. The slope of the line 7y = −2x + 3 is −2/7

We divide both sides of the equation by 7:

y = −2

7x +

3

7.

3. The slope of the line 9y − 4x + 7 = 0 is 4/9.

We subtract from both sides (−4x + 7). Next, we divide the equation by 9.

9y = 4x − 7,

y =4

9x − 7

9.

24

CHAPTER 3. ALGEBRA 3.4. GRAPHS

4. The slope of the line 3(y − 4) = 5(x + 7) is 5/3.

We divide by 3. Then, we add 4 to both sides and open the parenthesis.

y − 4 + 4 =5

3(x + 7) + 4,

y =5

3x +

35

3+ 4,

5

3x +

47

3.

5. The slope of the line 12y + 4x = 7 is −1/3.

We solve for y:

12y = −4x + 7,

y = − 4

12x +

7

12,

y = −1

3x +

7

12.

25


3.4.2 BAc3: Find the slope of a line from the coordinates of two points.

We usually denote the slope with m. If we have to denote more than one slope, we coulduse a subscript (example: m1). We use the formula

slope = m =yR − yQ

xR − xQ

.

where yQ means the y coordinate of Q and all other accordingly. In the formula, we canswap the place of the points P and Q (check!).

1. The coordinates of Q are (−3, 4) and the coordinates of R are (4, 7). The slope of theline that passes through Q and R

is

m =yR − yQ

xR − xQ

,

m =7 − 4

4 − (−3),

m =3

7.

2. The coordinates of Q are (−3, 4) and the coordinates of R are (4, 6). The slope of theline that passes through Q and R is

m =6 − 4

4 − (−3),

m =2

7.

3. The coordinates of R are (−3, 4/3) and the coordinates of S are (4, 2/3). The slope ofthe line that passes through R and S is

mRS =2/3 − 4/3

4 − (−3),

mRS = −2/3

7= −2

3× 1

7,

mRS = − 2

21.

26


4. The coordinates of S are (5/7, 4) and the coordinates of T are (4,−6). The slope ofthe line that passes through S and T is

mST =−6 − 4

4 − 5/7= (−10) ÷ 23

7,

mST = (−10) × 23

7,

mST = −70

23.

5. The coordinates of T are (−3,−4) and the coordinates of U are (4,−4). The slope ofthe line that passes through T and U is

mTU =(−4) − (−4)

4 − (−3),

mTU = 0.

We see that if the y coordinates are the same, the slope is zero.

Note if the x coordinates of the two points are the same, then the slope is undefined!This is because division by zero cannot be defined.

27


3.4.3 BAc4: Find the equation of a line from the coordinates of two

points.

Given the coordinates of two points, write an equation for the line in both point-slopeand slope intercept form if the line has a slope. If the line does not have a slope, write anequation in the form y = C.

We have a two phase process.First, we find the slope of the line. We can use the formulas from the previous section.Second, we write point-slope form of the equation. It is

y − y0 = m(x − x0).

We denote here y0 and x0 corresponding coordinates of either of the given points. To findintercept-slope form, we solve for y. We could open the parenthesis and the final form is

y = mx − mx0 + y0 = mx + (y0 − mx0).

1. The coordinates of Q are (−3, 4) and the coordinates of R are (4, 7). Equations of theline through Q and R are:For point-slope form, we first compute the slope:

m =7 − 4

4 − (−3)=

3

7.

Now, we use the formula with point Q (−3, 4):

y − 4 = m[x − (−3)],

y − 4 =3

7(x + 3).

For finding intercept-slope form, we solve for y:

y =3

7x +

(

3 × 3

7

)

+ 4,

y =3

7x +

37

7.

28


2. The coordinates of Q are (−3, 4) and the coordinates of R are (4, 6). Equations of theline through Q and R are:

m =6 − 4

4 − (−3)=

2

7.

We use point R (4,6) for point slope form:

y − 6 =2

7[x − (−3)],

y − 6 =2

7(x + 3).

The intercept-slope form is:

y =2

7x +

6

7+ 6,

y =2

7x +

48

7.

3. The coordinates of R are (−3, 4/3) and the coordinates of S are

(4, 2/3).

Equations of the line through R and S are:

m = 2/3−4/34−(−3)

= − 221

.

The point slope form with S (4, 2/3):

y − 2

3= − 2

21(x − 4).

The intercept-slope form:

y = − 2

21x +

8

21+

2

3,

y = − 2

21x +

22

21.

4. The coordinates of T are (−3, 4) and the coordinates of U are(−3,−4). The first coordinates are equal,therefore the slope is undefined. The linehas equation

x = −3.

5. The coordinates of S are (5/7, 4) and the coordinates of T are (4,−6). The equationsof the line through S and T are:

m =−6 − 4

4 − 5/7= −70

23.

The point slope form with S (4, 2/3)

y − (−6) = −70

23(x − 4).

29


The intercept-slope form:

y = −70

23x +

280

23− 6,

y = −70

23x +

142

23.

30


3.4.4 BAc5: Find the equation of a line given 1 point and the slope.

Given the slope of a line and the coordinates of a point on the line, write an equation forthe line in both point-slope and slope intercept form.Recall the formulas for the point-slope form and the intercept-slope form:

y − y0 = m(x − x0),

y = mx − mx0 + y0.

In the preceding, (x0, y0) are the coordinates of the given point, the first equation is point-slope form, the second is the intercept-slope form.

1. A line passes through (3,−4) and has slope −3. The point-slope equation is

y − (−4) = −3(x − 3).

The intercept-slope form of the equation is

y = −3x + 9 − 4,

y = −3x + 5.

2. A line passes through (7,−4) and has slope −3/7. The point-slope equation is

y − (−4) = −3

7(x − 7).


y = −3

7x + 3 − 4,

y = −3

7x − 1.

3. A line passes through (6,−11) and has slope −1/3. The point-slope equation is

y − (−11) = −1

3(x − 6).


y = −1

3x + 2 − 11,

y = −1

3x − 9.

31


4. A line passes through (2/5, 6) and has slope 5. The point-slope equation is

y − 6 = 5(x − 2

5).


y − 6 = 5x − 2,

y = 5x + 4.

5. A line passes through (3, 4/9) and has slope 5/9. The point-slope equation is

y − 4

9=

5

9(x − 3).


y =5

9x − 15

9+

4

9,

y =5

9x − 11

9.

32


3.4.5 BAc6: Determine and/or plot the intercepts of a line, given an

equation.

Given an equation for a line, give the horizontal and vertical intercepts of the line.

To find the intercept means we need to find points such that they are on the line, andone of them has first coordinate zero (0, y1) and the other one has second coordinatezero(x1, 0). In practice, we replace y with zero and solve for x - x-intercept and x withzero and solve for y -this is y-intercept.

(0,0)

(0,y )

(x ,0)

x-intercept

y-intercept

We sety=0 andsolve for x

We setx=0 andsolve for y ax+by+c=0

1

1

The other way is to plot the line, and then determine the intercepts. For this, we chooseany two x coordinates, plug then into the equation of the line and solve for y. We plot thepoints, and find the horizontal and vertical intercepts.

1. An equation for a line is −3x + 4y = 12.

If y = 0, then

−3x + 4 × 0 = 12,

x =12

−3= −4.

The horizontal intercept is (−4, 0).

If x = 0, then

−3 × 0 + 4y = 12,

y = 3.

The vertical intercept is (0, 3).

33


2. An equation for a line is 5x + 4y = 20.

If y = 0, then5x + 4 × 0 = 20,

x = 4.

The horizontal intercept is (4, 0) .

If x = 0, then5 × 0 + 4y = 20,

y = 5.

The vertical intercept is (0, 5).

3. An equation for a line is −3x + 4y = 25.If y = 0, then

−3x + 4 × 0 = 25,

x =25

−3= −25

3.

The horizontal intercept is (−25/3, 0).

If x = 0, then−3 × 0 + 4y = 25,

y =25

4.

The vertical intercept is (0, 25/4).

4. An equation for a line is −3(x − 2) + 4(y − 7) = 12.If y = 0 then

−3(x − 2) + 4(0 − 7) = 12,

−3x + 6 − 28 = 12,

−3x − 22 = 12,

−3x = 34

x = −34

3.

The horizontal intercept is (−34/3, 0).

34


If x = 0, then−3(0 − 2) + 4(y − 7) = 12,

6 + 4y − 28 = 12,

4y − 22 = 12,

4y = 34,

y =34

4=

17

2.


5. An equation for a line is x + 4y = 13.

If y = 0, thenx + 4 × 0 = 13,

x = 13.

The horizontal intercept is (13, 0).

If x = 0, then0 + 4y = 13,

y =13

4.


35


3.4.6 BAc9: Given an equation determine if a point is above or below

the graph.

Given an equation for a curve in the Cartesian plane and the coordinates of a point in theplane, determine if the point is above or below the curve.Example: An equation for a line is 3x+4y = 12. The point with coordinates (4, 5) is abovethis line because the point (4, 0) is on the line and 0 < 5.

In general, we evaluate the function at the point x. We compare the value of the functionto the y coordinate of the point.Geometrically, we plot the function and a vertical line at horizontal coordinate x0. If com-puted value of the function is bigger than the second coordinate of the point, then thepoint is below the graph.

I will denote <>= the process of comparing of two numbers. The expression y(1)means that I evaluate the function y at the point when x = 1.

1. Is the point (0, 4) above, below, or on the curve y = x3 + 2x − 1?We compute the value of y at x = 0, and compare it with second coordinate of thepoint specified - 4.Is y(0) = 03 + 2 × 0 − 1 = −1 <>= 4?

y(0) = −1 < 4.

We see that y(0) is less than 4. It means that the point (0, 4) is above the graph.

2. Is the point (−3, 1) above, below, or on the curve y = −2x2 + 2x − 1?Is y(−3) = −2(−3)2 + 2(−3) − 1 = −18 − 6 − 1 <>= 1?

y(−3) < 1.

The point (−3, 1) is above the graph y = −2x2 + 2x − 1 because y(−3) = −25 < 1.

3. Is the point (1, 1) above, below, or on the curve y = x3 + 2x − 2?Is y(1) = 13 + 2 × 1 − 2 <>= 0?

y(1) = 1.

The point (1, 1) is on the graph y = x3 + 2x − 2 because y(1) = 1.

4. Is the point (3,−4) above, below, or on the curve y = x3 + 2x − 100?Is y(3) = 33 + 2 × 3 − 100 <>= −4?

y(3) < −4.

The point (3,−4) is above the graph because y(3) = −67 < −4.

36


5. Is the point (16, 50) above, below, or on the curve y = 5x + 2√

x − 1?Is y(16) = 5 × 16 + 2

√16 − 1 = 80 + 8 − 1 <>= 50?

y(16) > 50.

The point (16, 50) is below the graph y = 5x + 2√

x − 1 because y(16) = 87 > 50.

37

3.5. QUADRATIC/RATIONAL EQUATIONS CHAPTER 3. ALGEBRA

3.5 Quadratic/Rational Equations

3.5.1 BAd1: Quadratics of the form ax2 + b = 0 or (ax + b)2 + c = 0.

Solve a quadratic equation where the square has been completed.

It is much easier to solve a quadratic equation once the square has been completed. Wesimply move the free term on the other side and take square root from both sides of theequation. We have two equations with one and the same left side and opposite right sides.

1. Solve 4x2 − 36 = 0 for x. We add the free term 36 to both sides of the equation:

4x2 − 36 + 36 = 0 + 36.

We solve for x:

x2 =36

4= 9,

|x| =√

9 = 3.

Recall that when taking square root we compute the principal square root, i.e. thesquare root without its sign. There are two numbers with absolute value of 3: x1 = 3and x2 = −3. The solutions are: x1 = +3 and x2 = −3.

2. Solve 4(x − 12)2 − 36 = 0 for x.

4(x − 12)2 = 36,

(x − 12)2 =36

4= 9,

x − 12 = +√

9 = +3 or x − 12 = −√

9 = −3.

We solve these equations and find

x − 12 = 3,

x1 = 15.

x − 12 = −3,

x2 = −9.

The solutions are x1 = 15 and x2 = 9.

3. Solve −3a2 + 36 = 0 for a.−3a2 = −36,

a2 =−36

−3= 12,

38

CHAPTER 3. ALGEBRA 3.5. QUADRATIC/RATIONAL EQUATIONS

a1 = +√

12 = +2√

3,

a2 = −√

12 = −2√

3.

The solutions are a1 = +2√

3 and a2 = −2√

3. Note, that the solutions may not berational numbers so that the FOIL device is not always useful.

4. Solve 5(x + 3)2 − 25 = 0 for x.

5(x + 3)2 = 25,

(x + 3)2 =25

5= 5,

x + 3 = +√

5 or x + 3 = −√

5.

x1 = −3 +√

5,

x2 = −3 −√

5

The solutions are x1 = −3 +√

5 and x2 = −3 −√

5

5. Solve 2(x + 9)2 − 162 = 0 for x.

2(x + 9)2 = 162,

(x + 9)2 =162

2= 81,

x + 9 = +√

81 = +9 or x + 9 = −√

81 = −9.

x1 = −9 + 9 = 0,

x2 = −9 − 9 = −18.

The solutions are x1 = 0 and x2 = −18.

39


3.5.2 BAd2: Quadratics of the form ax2 + bx + c = 0 with a 6= 1 by

factoring over the integers.

Solve a quadratic equation where the quadratic expression factors over the integers.

To solve quadratic equation we use zero product principle: the product of two numbersis zero if and only if at least one of the factors is zero. For quadratics, to find the factorsof the quadratic expression and apply the principle means to make each factor equal to 0.If we have rational solutions to the equation, we can use FOIL method for factoring. Thedetails will be shown with #1.

Be advised this is a method of trials, so you must be consistent and try until you findthe answer or exhaust all possibilities. Check your work carefully to avoid making errors.

1. Solve 2x2 + 5x − 3 = 0 for x.FOIL method for factoring. We need two First terms with product (+2): the only com-bination is (+2, +1). We need two Last terms with product (−3): they could be(+1,−3) and (−1, +3). We compute the sum Outer terms product plus Inner termsproduct until we have a sum equal to (+5).

In the table, FTPr means First terms product, LTPr means Last terms product andOIpSum means Outer and Inner terms product sum. Outer terms are shown in outerpositions, and Inner terms in inner. The information is organized in the followingtable. For example the first row represents the quadratic in factors product [(+1)x+(+1)] × [(+2)x + (−3)] - the terms come in the order they are in FOIL.

FTPr=2 LTPr=-3 OIpSum=+5+1, +2 +1, -3 -1+1, +2 -1, +3 +1+2, +1 +1, -3 -5+2, +1 -1, +3 +5

We see we have the sum we need in the last row. Then, the equation factors

0 = 2x2 + 5x − 3 = (2x − 1)(x + 3).

We now can solve2x − 1 = 0 and x + 3 = 0.The first one gives the solution x = 1

2and second x = (−3).

For all the following problems, only the table, resulting equations and solutions willbe shown.

2. Solve 4x2 + 4x − 3 = 0 for x.

40


FTPr=4 LTPr=-3 OIpSum=+4+1, +4 +1, -3 +1+1, +4 -1, +3 -1+4, +1 +1, -3 -11+4, +1 -1, +3 +11+2, +2 +1, -3 -4+2, +2 -1, +3 +4

The equivalent forms of the equation are

0 = 4x2 + 4x − 3 = (2x − 1)(2x + 3).

The solutions are x = 12

and x = −32.

3. Solve 6a2 + 13a + 6 = 0 for a.

FTPr=6 LTPr=6 OIpSum=+13+1, +6 +2, +3 +15+1, +6 +3, +2 +20+1, +6 +6, +1 +37+1, +6 +1, +6 +12+2, +3 +1, +6 +15+2, +3 +3, +2 +13

The equivalent forms of the equation are

0 = 6a2 + 13a + 6 = (2a + 3)(3a + 2).

The solutions of the equation are a1 = −32

and a2 = −23.

4. Solve 6x2 − x − 15 = 0 for x.

FTPr=6 LTPr=-15 OIpSum=-1+1, +6 -5, +3 -27+1, +6 +3, -5 +13+2, +3 -5, +3 -11+2, +3 +3, -5 -1

The equation is0 = 6x2 − x − 15 = (2x + 3)(3x − 5).

The solutions of the equation are x1 = −32

and x2 = 53.

41


5. Solve 6x2 + 13x − 15 = 0 for x. We can use the table above to solve # 5 as well. Thetable is reduced to only one row:

FTPr=6 LTPr=-15 OIpSum=+13+1, +6 +3, -5 +13

The equation is0 = 6x2 + 13x − 15 = (x + 3)(6x − 5).

The solutions of the equation are x1 = −3 and x2 = 56.

42


3.5.3 BAd3: Quadratics using the quadratic formula, where the dis-

criminant might be positive, negative or zero.

Solve a quadratic equation where the roots may not be rational numbers. Either we com-plete the square or use the quadratic formula.We use the quadratic formula. It can be used in every case. If the equation is

ax2 + bx + c = 0

and we solve for x, the solutions of the equation are

x1,2 =−b ±

√b2 − 4ac

2a.

This means there are two solutions:

x1 =−b +

√b2 − 4ac

2a

and

x2 =−b −

√b2 − 4ac

2a.

The other general way for solving quadratic equations is by completing the square. Weilustrate it in the following problem.

1. 2x2 + 5x − 4 = 0.We add 4 to both sides of the equation.

2x2 + 5x = 4.

We divide both sides by 2.

x2 +5

2x = 2.

Here is why this technique is called ”completing the square”. On the left side wewant to have a perfect square. The first term is x. We know the crossproduct is 5

2x.

This is the double product of the first and the second term. We know that the firstone is x and determine the second one (called below b):

2 × x × b =5

2x.

We solve for b: b = 54. Therefore we need to add to both sides the square of b,i.e. -

(54)2:

x2 + 2 × 5

4x +

(

5

4

)2

= 2 +

(

5

4

)2

.

43


We factor the square on the left.

(

x +5

4

)2

=57

16.

We take square roots from both sides and have:

x +5

4= +

√

57

16

or

x +5

4= −

√

57

16.

We have plus or minus because there are two numbers with square√

57/16, just as(+3)2 = (−3)2 = 9. We solve for x:

x1 = −5

4+

√57

4,

x2 = −5

4−

√57

4.

2. 4x2 − 4x − 5 = 0.We will use the same method for this equation.

4x2 − 4x = 5,

(2x)2 − 2(2x) × 1 + 12 = 5 + 1,

(2x + 1)2 = 6.

2x + 1 = +√

6

or2x + 1 = −

√6.

The solutions are x1 = (−1 +√

6)/2 and x2 = (−1 −√

6)/2.

3. 6a2 + 13a + 7 = 0:Here we will use the quadratic formula. Since a = 6, b = 13 and c = 7 we use theformulas:

x1 =−b +

√b2 − 4ac

2a

and

x2 =−b −

√b2 − 4ac

2a.

x1 =−(+13) +

√

(+13)2 − 4 × (+6) × (+7)

2 × 6,

44


x1 =−13 ±

√169 − 168

12,

x1 =−13 + 1

12= −1.

For x2 we have

x2 =−13 − 1

12= −7

6.

Then, the solutions are x1 = −1 and x2 = −7/6.

4. x2 + 2x + 15 = 0.

x1 =−2 + 22 − 4 × 1 × 15

2 × 1,

x1 =−2 +

√−56

2.

Recall the fact that√−1 = i is an imaginary number which we assume has square

equal to (−1). We can write then√−56 =

√

(−1)56 =√−1

√56 =

√56i.

x1 =−2 +

√56i

2.

We can factor out 2:−2 +

√56i

2=

−2 −√

22 × 14i

2

=2(−1 +

√14i

2= −1 +

√14i.

so the solution isx1 = −1 +

√14.

The same way

x2 = −1 −√

14i.

The solutions are x1 = −1 + i√

14 and x2 = −1 − i√

14.

5. 6x2 + 13x + 15 = 0.

x1 =−13 +

√132 − 4 × 6 × 15

2 × 6,

x1 =−13 +

√−191

12,

x1 =−13 + i

√191

12.

The same way:

x2 =−13 − i

√191

12.

The solutions are x1 = (−13 + i√

191)/12 and x2 = (−13 − i√

191)/12.

45


3.5.4 BAd4: Equations quadratic in u, where u = x2, x3, ....

Solve an equation which is quadratic in a power of x.

To solve an equation which is quadratic in a power of x, we make use of the property ofexponents as in example shown below:

(x2)3 = (x3)2 = x2×3 = x6.

The property states if we have a power of a variable (say x), we can break the exponentdown into two factors and when computing this way, the result is the same. You cancheck for yourself (22)3 = (23)2 = 26.We can use this rule for rational exponents as well. Recall the denominator of the expo-nent is the root which we take to compute the number.

1. x2/3 + 3x1/3 + 2 = 0.We see that 2

3= 2 × 1

3. We can simplify the equation by letting

x1/3 = u.

This is our substitution. Then, we have

u2 = x2/3.

The equation becomesu2 + 3u + 2 = 0.

We see we can solve this equation easily (u+1)(u+2) = 0. The solutions are u1 = −1and u2 = −2.The equation is not solved yet! We need to go back and use the substitution to find x:

x1/3 = −1.

We findx1 = −1.

The other solution u2 = −2 will give us

x1

3 = −2,

x2 = −8.

2. x6 + 3x3 + 2 = 0. We substitutex3 = v.

The equation becomes

x6 + 3x3 + 2 = (x3)2 + 3(x3) + 2 = v2 + 3v + 2 = 0.

46


The last right equation is solved above. We have the solutions v1 = −1 and v2 = −2.We now solve

x3 = −1,

x3 + 1 = 0.

We factor this equation and find (x + 1)(x2 − x + 1) = 0. Now, we use zero productprinciple and find

x + 1 = 0,

x1 = −1.

Next, we solvex2 − x + 1 = 0.

Using the quadratic formula, we find two complex number roots:

x2 =1 +

√

(−1)2 − 4 × 1 × 1

2,

x2 =1

2+ i

√3

2.

The other root is

x3 =1

2− i

√3

2.

For the second solution of u we solve the same way:

x3 = −2,

x3 + 2 = 0,

(x +3√

2)(x2 − 3√

2x + (3√

2)2) = 0.

This equation has three other roots:

x4 = − 3√

2.

To ease the computations, we first compute the discriminant of (x2− 3√

2x+( 3√

2)2) =0:

D = (− 3√

2)2 − 4 × 1 × (3√

2)2) = −3 3

√

(22).

The remaining solutions are

x5 =

3√

2 +√

−3 3

√

(22)

2,

x5 =3√

2

2+ i

3√

2

√3

2and

x6 =3√

2

2− i

3√

2

√3

2.

All the solutions of the equation x6 + 3x3 + 2 = 0 are x1 = −1, x2 = 1/2 + i√

3/2,x3 = 1/2 − i

√3/2, x4 = − 3

√2, x5 = 3

√2(1/2 + i

√3/2) and x6 = 3

√2(1/2 − i

√3/2).

47


3. 2x4 − 3x2 + 1 = 0.We substitute x2 = u.

2u2 − 3u + 1 = 0.

We use FOIL and find (2u − 1)(u − 1) = 0. Thus, u1 = 12

and u2 = 1. We find all thesolutions for x now:

x2 =1

2,

x1,2 = ±√

1

2= ±

√2

2.

The first two solutions are x1 =√

2/2 and x2 = −√

2/2. Finally, we solve

x2 = 1,

x3,4 = ±1.

The other two solutions are x3 = 1 and x4 = −1.

4. x + 5 + 4x−1 = 0.This equation is not in a quadratic form for any power of x. But, we can transformit in this form by multiplying it by x:

(x + 5 + 4x−1)x = 0 × x

x2 + 5x + 4 = 0

This is a regular quadratic equation and we solve it for x:

(x + 4)(x + 1) = 0

The solutions are x1 = −4 and x2 = −1.

5. x2 − 2 + x−2 = 0.We use the same idea for this equation (but multiplying by x2):

x4 − 2x2 + 1 = 0

We substitute x2 = u:u2 − 2u + 1 = 0

This equation is (u− 1)2 = 0. Therefore, there is only one repeated solution u1,2 = 1.We solve for x now and find x1 = 1 and x2 = −1. The original equation is not apolynomial equation so that it has only two solutions as stated above.

48


3.5.5 BAd5: Rational equations leading to quadratic equations.

Solve a rational equation which can be reduced to a quadratic equation.

We need to transform the rational equation to quadratic equation by using a processknown as ”clearing fractions”.We multiply the equation by the Least Common Denominator of all rational fractions in theequation. This is not all. We need to list all restrictions, i.e., all values of x for which theequation is defined. Recall we cannot divide by zero.

1.2x − 5

x + 1= x + 2.

We see that the LCD = x + 1. This denominator must not be equal to zero, so weneed x 6= −1. We multiply now the equation by (x + 1):

2x − 5 = (x + 2)(x + 1).

We simplify it and solve for x:

2x − 5 = x2 + 3x + 2,

x2 + x + 7 = 0.

We use the quadratic formula:

x1 =−1 +

√12 − 4 × 1 × 7

2

and

x2 =−1 −

√12 − 4 × 1 × 7

2.

There are two complex number solutions:

x1 = −1

2+ i

3√

3

2

and

x2 = −1

2− i

3√

3

2.

2.2x − 5

x + 1=

x + 2

x − 1.

LCD = (x + 1)(x − 1).We restrict x:x 6= −1 and x 6= +1.We can clear the fractions now:

(2x − 5)(x − 1) = (x + 2)(x + 1),

49


2x2 − 7x + 5 = x2 + 3x + 2,

x2 − 10x + 3 = 0.

We use quadratic formula:

x1 =10 +

√

(−10)2 − 4 × 1 × 3

2

and

x2 =10 −

√

(−10)2 − 4 × 1 × 3

2.

The solutions arex1 = 5 +

√22

andx2 = 5 −

√22.

3.2x − 5

x + 1= 3x + 2.

LCD = x + 1, then x 6= −1.Clearing fractions:

2x − 5 = (3x + 2)(x + 1),

2x − 5 = 3x2 + 5x + 2,

3x2 + 3x + 7 = 0.

The solutions are

x1 =−3 +

√32 − 4 × 3 × 7

2 × 3

and

x2 =−3 −

√32 − 4 × 3 × 7

2 × 3,

i.e.

x1 = −1

2+ i

√75

6

and

x2 = −1

2− i

√75

6.

4.1

x + 1+

1

x − 1= 3.

LCD = (x + 1)(x − 1), thus x 6= −1 and x 6= +1.Clearing fractions:

[

1

x + 1+

1

x − 1

]

(x + 1)(x − 1) = 3(x + 1)(x − 1).

50


We open the parenthesis on the left and remove the common factors:

(x − 1) + (x + 1) = 3(x2 − 1).

We solve for x:3x2 − 3 = 2x,

3x2 − 2x − 3 = 0.

The solutions are:

x1 =2 +

√

(−2)2 − 4 × 3 × (−3)

2 × 3

and

x1 =2 −

√

(−2)2 − 4 × 3 × (−3)

2 × 3,

i.e.

x1 =1

3+

√10

3

and

x2 =1

3−

√10

3.

5. Solve3x − 5

x − 1= 2x − 1.

LCD = x − 1, thus x 6= 1.

Clearing fractions:3x − 5 = (2x − 1)(x − 1),

3x − 5 = 2x2 − 3x + 1,

2x2 − 6x + 6 = 0,

x2 − 3x + 3 = 0.

The solutions are:

x1 =3 +

√

(−3)2 − 4 × 1 × 3

2 × 1

and

x2 =3 −

√

(−3)2 − 4 × 1 × 3

2 × 1,

i.e.

x1 =3

2+ i

√3

2

and

x2 =3

2− i

√3

2.

51


3.5.6 BAd6: Rational equations leading to linear equations.

Solve a rational equation that reduces to a linear equation.There is no difference between solving rational equations leading to quadratic equationsand those leading to linear equations. We simply have only one solution.

1.2x2 + 5

2x + 1= x + 2.

LCD = 2x + 1, thus x 6= −1/2.Clearing fractions:

2x2 + 5 = (x + 2)(2x + 1),

2x2 + 5 = 2x2 + 5x + 2.

We solve for x:

5x = 3,

x =3

5.

2.2x − 5

2x + 1=

x + 2

x − 1.

LCD = (2x + 1)(x − 1), thus x 6= −1/2 and x 6= 1.Clearing fractions and solving:

(2x − 5)(x − 1) = (2x + 1)(x + 2),

2x2 − 7x + 5 = 2x2 + 5x + 2,

−12x = −3,

x =1

4.

3.3x2 − 5

x + 1= 3x + 2.

LCD = x + 1, thus x 6= −1.Clearing fractions and solving:

3x2 − 5 = (3x + 2)(x + 1),

3x2 − 5 = 3x2 + 5x + 2,

5x = −7,

x = −7

5.

52


4.x

x + 1+

1

x − 1= 1.

LCD = (x + 1)(x − 1), thus x 6= ±1.We clear fractions and solve:

[

x

x + 1+

1

x − 1

]

(x + 1)(x − 1) = 1 × (x + 1)(x − 1),

x(x − 1) + 1 × (x + 1) = x2 − 1,

x2 − x + x + 1 = x2 − 1.

We have now+1 = −1 FALSE!

We see this equation does not have any solution.

5.4x2 − 5x + 7

2x − 1= 2x − 1.

LCD = 2x − 1, thus x 6= 1/2.Clearing fractions and solving:

4x2 − 5x + 7 = (2x − 1)(2x − 1),

4x2 − 5x + 7 = 4x2 − 4x + 1,

−x = −6,

x = 6.

53


3.5.7 BAd7: Solve literal equations involving rational expressions.

Solve a rational equation in several variables for a specific variable.

We use the same method when solving literal equations as when we use numbers. Theonly difference is we keep the expression with all variables. Keep in mind variables rep-resent numbers and, if we cannot do something with numbers, we cannot do it withvariables too (example: dividing by zero is not defined so we cannot allow to have de-nominators equal to zero).

Solve each of the following equations for b.

1.1

b+

1

a= 1.

We isolate on the first step 1/b by subtracting 1/a from both sides and simplify:

1

b= 1 − 1

a=

a − 1

a.

We take the reciprocal of both sides (after restricting a 6= 1 since 1/b cannot be zero):

b =a

a − 1.

2.a

b+

c

d= 1.

To isolate b on one side, we first subtract the fraction c/d from both sides and sim-plify:

a

b= 1 − c

d=

d − c

d.

We multiply both sides bydb

d − c( after restricting c 6= d):

a

b× b

d

c − d=

c − d

d× d

c − db.

Dividing out common factors will give us

b =ad

d − c.

3.1

b+

1

a=

1

a + b.

We can clear fractions as we have it done before, restricting a 6= 0 b 6= 0 and a 6= −b.We can multiply by ab(a + b):

a(a + b) + b(a + b) = ab,

54


a2 + ab + ab + b2 = ab.

We subtract ab from both sides of the equation and see that we have a quadraticequation for b. We solve it for b using quadratic formula:

b2 + ab + a2 = 0,

b1,2 =−a ±

√a2 − 4 × 1 × a2

2 × 1.

Then,

b1 = −a

2+ i

√3|a|2

,

b2 = −a

2− i

√3|a|2

.

4.a

b+

b

a= 1.

We clear fractions after restricting a 6= 0 and b 6= 0:

a

bab +

b

a= 1ab,

a2 + b2 = ab.

We subtract ab from both sides of the equation. This is a quadratic equation for b.We solve it using quadratic formula:

b2 − ab + a2 = 0,

b1,2 =−(−a) ±

√

(−a)2 − 4 × 1 × a2

2 × 1.

Then,

b1 =a

2+ i

√3|a|2

,

b2 =a

2− i

√3|a|2

.

5.1

b− 1

a= 2.

The restrictions are a 6= 0 and b 6= 0. We can proceed as in #1: we isolate 1/b on oneside of the equation, and simplify.

1

b= 2 +

1

a=

2a + 1

a.

We cannot have the left side zero, so we restrict the right side not to be equal to zeroa 6= −1/2. We then take the reciprocal of both sides:

b =a

2a + 1.

55

3.6. RADICALS AND FRACTIONIAL EXPONENTS CHAPTER 3. ALGEBRA

3.6 Radicals and Fractionial exponents

3.6.1 BBa4: Express radicals in terms of fractional exponents and vice

versa.

Rewrite each expression using the alternate form for expressing roots. Do not simplify.

Recall that A1/n is actually n

√A and ( n

√B)m is Bm/n:

A1/n =n

√A

and

(n

√B)m = Bm/n.

1.√

36m = 1 and n = 2.

= 361/2.

2. 3√−27

m = 1 and n = 3.= (−27)1/3.

3. 1003/2

m = 3 and n = 2.= (

√100)3.

4. 255/2

n = 2 and m = 1.= (

√25)5.

5. (−32)2/5

m = 2 and n = 5.= ( 5

√−32)2.

56

CHAPTER 3. ALGEBRA 3.6. RADICALS AND FRACTIONIAL EXPONENTS

3.6.2 BBa1: Evaluate numerical expressions involving radicals and/or

fractional exponents.

Simplify numerical expressions involving roots in either radical or fractional exponentform.

1.√

36

= 6.

2. 3√−27

= −3.

3. 1003/2

= (√

100)3 = 103 = 1000.

4. 255/2

= (√

25)5 = 55 = 3125.

5. (−32)2/5

= ( 5√−32)2 = (−2)2 = 4.

57


3.6.3 BBa2: Simplify radical expressions.

Reduce and combine radicals in each radical expression. You may assume all variablesrepresent positive quantities.Recall a perfect square is a number or expression which is a square of a rational numberor another expression. Example 4 = 22 and a4 = (a2)2. The same way we can defineperfect third power, fourth power and so on. In general, perfect n-th power is a numberor expression which is an n-th power of a number or expression.

1. 3

√

16x5y3/z6

We see that 16 = 23 × 2 and x5 = x3x2. The other expressions y3 and z6 are perfectthird powers (also known as perfect cube).

= 3

√

(23x3y3/z3)(2x2)

= 3

√

(2xy/z2)3 3√

2x2

= (2xy

z2)3/3 3

√2x2 =

2xy

z2

3√

2x2.

2. 4

√

81x5y9/z6

= 4

√

(34x4y8/z4)(xy/z2)

= [(3xy2)4]1/4 4

√

xy/z2

=3xy2

z24

√

xy

z2.

3.√

45 +√

80 −√

20=

√32 × 5 +

√42 × 5 −

√22 × 5

= 3√

5 + 4√

5 − 2√

5 = 5√

5.

4. 3

√

27x7y3/z8

= 3

√

(33x6y3/z6)(x/z2)

= [(3x2y/z2)3]1/3 3

√

x/z2

=3x2y

z23

√

x

z2.

5.√

12x4y3

=√

(22x4y2)(3y)

=[

(2x4y)2]1/2 √

3y

= 2x2y√

3y.

58


3.6.4 BBa3: Rationalize the numerator or denominator of a quotient in-

volving radicals.

To rationalize the numerator or denominator of an expression, we use the following for-mula:

(A + B)(A − B) = A2 − B2.

We will multiply by an expression of one, containing the conjugate of the radical number.The conjugate of A + B is A − B, and the conjugate of A − B is A + B.

1.

√x −√

y

(x − y)The conjugate of (

√x − √

y) is (√

x +√

y). Therefore, we need to multiply with ex-

pression of one√

x+√

y√

x+√

y:

=

√x −√

y

x − y×

√x +

√y√

x +√

y

=(√

x)2 − (√

y)2

(x − y)(√

x +√

y)

=x − y

x − y× 1√

x +√

y=

1√x +

√y

, if x 6= y.

2.

√x +

√y

x + yThe conjugate is

√x −√

y.

=

√x +

√y

x + y×

√x −√

y√x −√

y

=(x − y)

(x + y)(√

x −√y)

, if x 6= y.

3.

√x + 2 −

√x − 2

4The conjugate is

√x + 2 +

√x − 2.

=

√x + 2 −

√x − 2

4×

√x + 2 +

√x − 2√

x + 2 +√

x − 2

=(√

x + 2)2 − (√

x − 2)2

4(√

x + 2 +√

x − 2)

=x + 2 − x + 2

4(√

x + 2 +√

x − 2)

=1√

x + 2 +√

x − 2.

59


4.3x√

x2 + 1 −√y

We need now the conjugate of the denominator:The conjugate is

√x2 + 1 +

√y.

=3x√

x2 + 1 −√y×

√x2 + 1 +

√y√

x2 + 1 +√

y

=3x(

√x2 + 1 +

√y)

x2 + 1 − yif x2 + 1 6= y.

5.5√

x −√y

The conjugate is√

x +√

y.

=5√

x −√y×

√x +

√y√

x +√

y

=5(√

x +√

y)

x − y, if x 6= y.

60


3.6.5 BBa5: Multiply radical expressions (monomials, binomials).

Multiply the radical expression. You may assume all variables represent positive quanti-ties. Combine like terms, but do not simplify otherwise.

1.√

2x + 3y√

3

=√

3(2x + 3y).

2. 4

√

81x5y9 4√

7xyz

= 4

√

(81x5y9)(7xyz) = 4

√

567x6y10z.

3.√

45√

10=

√45 × 10 =

√450.

4. 3

√

27x7y3 3√

x + y

= 3

√

(27x7y3)(x + y).

5. 3

√

12x4y3 3

√

5x2y3

= 3

√

(12x4y3)(5x2y3).

61


3.6.6 BBa6: Simplify and combine like radicals.

Simplify the radical expression. You may assume all variables represent positive quanti-ties. Combine like terms and extract all possible roots.Recall that a square root of a number, is a number which square is the given number. Wecall a number or expression a perfect square if it is a square of another rational number orexpression. We simplify by extracting the largest possible perfect square.The same way is defined perfect cube — third power of a number, perfect forth and soon...

1.√

54√

3=

√54 × 3 =

√2 × 33 × 3

=√

81 × 2 =√

81√

2 = 9√

2.

2. 4

√

81x5y9 4

√

7xy2z

= 4

√

(81x4y8)(7x2y3z)

= [(34xy2)4]1

44

√

7x2y3z = 3xy2 4

√

7x2y3z.

3.√

45√

20=

√32 × 5 × 5 × 22 =

√32 × 52 × 22

=√

302 = 30.

4. 3

√

27x7y3 3√

3x + 6y

= 3

√

(33x6y3)x(3x + 6y)

=[

(3x2y)3]1/3 3

√

x(3x + 6y)

= 3x2y 3

√

x(3x + 6y).

5. 3

√

12x4y5 3

√

5x2y4

= 3

√

60x4+2y5+4

=[

(x2y3)3]1/3 3

√60 = x2y3 3

√60.

62


3.6.7 BBa7: Simplify expressions involving radicals by changing to frac-

tional exponents and using exponent laws.

Use fractional exponents to combine like terms.Recall the index of the radical is the denominator of the exponent fraction. We multiplyexponents with one and the same base by adding the exponents. We use the same rulesfor adding fractions as usual.

1. 4

√

x5y 3√

xy

= (x5y)1/4(xy)1/3 = x5/4+1/3y1/4+1/3

= x19/12y7/12.

2. 3

√

x4y2 2

√

2xy3

= (x4y2)1/3(2xy3)1/2

= x4/3+1/2 y2/3+3/221/2 = x11/6 y13/6 21/2.

3. 5

√

x4y2 3

√

x3y5

= (x4y2)1/5 (x3y5)1/3

= x4/5+1 y2/5+5/3 = x9/5 y31/15.

4. 2

√

x4y2 4

√

2xy3

= (x4y2)1/2 (2xy3)1/4

= 21/4x2+1/4 y1+3/4 = 21/4x9/4y7/4.

5. 3

√

x5y4 6

√

x5y= (x5y4)1/3(x5y)1/6

= x5/3+5/6 y4/3+1/6 = x5/2y3/2.

63

3.7. NEGATIVE EXPONENTS CHAPTER 3. ALGEBRA

3.7 Negative Exponents

3.7.1 BBb2: Use reciprocal rule to eliminate denominator negative ex-

ponents.

Write each expression so it has no negative exponents.

Recall for any real number a and and any integer n we have

a−n =1

an.

Examples:

3−2 =1

32

and1

5−3= 53.

It turns out removing the negativity of the exponent means to raise to the opposite posi-tive exponent the reciprocal of the number-base.All the laws of exponents hold when using negative exponents.

1.x3y4

x−3y−1

= x3y4(x3y1) = x3+3y4+1 = x6y5.

We see, in this example, that the reciprocal of a number could be expressed as thenumber raised to power (−1).

2.6x2y−5

3x−3y−7

= 2x2y−5(x3y7) = 2x2+3y−5+7 = 2x5y2.

We see the laws of exponents hold when the exponents are negative.

3.w2y3

x−4y−2

= w2y3(x4y2) = x4y5w2.

4.x3/2y4

x−1/2y−2

= x3/2y4(x1/2y2) = x(3/2)+(1/2)y6 = x2y6.

Note the laws hold even when the exponents are fractions.

64

CHAPTER 3. ALGEBRA 3.7. NEGATIVE EXPONENTS

5.x−3/2y1/2

x−3y−1

= x−3/2y1/2(x3y) = x(−3/2)+3y(1/2)+1 = x3/2y3/2.

In conclusion, example #2 underlines the use of exponent laws with negative numbersand #4 — fractional exponents. In all cases, when the base is the same, we multiply byadding exponents and divide by subtracting the exponents.

65


3.7.2 BBb3: Simplify negative powers of sums and differences.

Write each expression so that it has no negative exponents.

We will see the base could not only be a number but it could be an expression contain-ing addition and subtraction.To remove the negativity of the base, we exchange the base with its reciprocal. Example:

4−2 =

(

1

4

)2

=1

42.

1. (x + 2)−1

=1

x + 2.

2. (x − 2)−2

=1

(x − 2)2.

3. (x + y + 2)−2

=1

(x + y + 2)2.

4. (3x + 2)−2 =

=1

(3x + 2)2.

5. (x − 1)−3

=1

(x − 1)3.

66

CHAPTER 3. ALGEBRA 3.7. NEGATIVE EXPONENTS

3.7.3 BBb4: Evaluate or simplify expressions involving negative expo-

nents.

1. If a = 2 and b = 5 then a−2 + b

= 2−2 + 5 =1

22+ 5 = 5

1

4=

21

4.

2. If a = −2 and b = 5 then a−3 + b−1

= (−2)−3 + 5−1 =1

(−2)3+

1

51

= −1

8+

1

5=

−5 + 8

8 × 5=

3

40.

3. Combine like terms and eliminate the negative exponents:

a + b

a−1 + b−1

=a + b

1/a + 1/b

=a + b

(b + a)/(ab)

Recall that dividing a number to another number is actually multiplication with thereciprocal of the second number:

= (a + b) × ab

a + b= ab, if a 6= −b.

67



2a2 + b

a−2 + b−1

We use the same method:

=2a2 + b

1/a2 + 1/b

=2a2 + b

(b + a2)/(a2b)

=(2a2 + b)a2b

a2 + b.


2a2b

a−2 + b−3

=2a2b

1/a2 + 1/b3

=2a2b

(b3 + a2)/(a2b3)

=(2a2b)(a2b3)

b3 + a2=

2a4b4

a2 + b3.

68

CHAPTER 3. ALGEBRA 3.8. ABSOLUTE VALUE

3.8 Absolute Value

3.8.1 BBc1: Evaluate numerical expressions involving absolute values.

Give the value of an expression involving absolute value.Example: If x = 2 then |x − 1| − |3 − x| = 0.Recall the absolute value of a number is the number without its sign. So, | + 2| = 2 and| − 3| = 3.

1. If x = 5, then |3x − x2| − |3 − 2x|

= |3 × 5 − 52| − |3 − 2 × 5|

= | − 10| − | − 7|= 10 − 7 = 3.

2. If x = −5, then −|x − x2| + |3 + 2x|

= −|(−5) − (−5)2| + |3 + 2 × (−5)|

= −| − 5 − 25| + |3 − 10|= −30 + 7 = −23.

3. If x = 0, then |3x − x2| − | − 3 − 2x|

= |3 × 0 − 02| − | − 3 − 2 × 0|

= 0 − 3 = −3.

4. If x = −3, then√

x2 − |x|=

√

(−3)2 − | − 3|Recall the square root is only the positive square root of a number, so

√

(−3)2 =√9 = 3, therefore

= 3 − 3 = 0.

5. If x = −1, then |3 − 4x2| + |8 − 2x| + x

= |3 − 4(−1)2| + |8 − 2(−1)| + (−1)

= |3 − 4| + |8 + 2| − 1 = 1 + 10 − 1 = 10.

69

3.8. ABSOLUTE VALUE CHAPTER 3. ALGEBRA

3.8.2 BBc2: Recognize the graph of y = |x| and its variations.

Example: Which of the following is the graph of y = |x| + 1?

EXAMPLE EXPLANATION: We can consider the most basic absolute value graph.What we need is another expression, whose graph we know. This is the linear expression.The absolute value is the number without its sign. We can define:|x| = −x when x < 0

When x is negative,then absolute value ofx is the opposite: -x

|-3|= -(-3)=3y = |x|

and

|x| = +x when x ≥ 0.

When x is positive,absolute value of x isthe number itself.

|2| = 2

y= |x

|

Both are parts of a line but not the whole line. The graph of absolute value function isgiven below.It appears as the latin letter V. All functions with absolute value have this basic shape.The other characteristics of this graph are its vertex and axis of symmetry. The vertex forthis graph is point (0, 0) and axis of symmetry is x = 0.

70


Basic shape of y = |x|

a) b) c) d)

–10

–8

–6

–4

–20

2

4

6

8

10

y

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

Only graph b) has desired shape.

Solution: b

1. Circle the graph of y = |3x| + 1.

a) b) c) d)

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

y

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

There are three graphs with desired shape. We consider a), c) and d). Note the func-tion is symmetric about x = 0. This means if we replace a number, say x = 2, withits opposite, x = −2 the function value is the same. Indeed, 6 = |3 × 2| = f(2) and6 = |3 × (−2)| = f(−2)|. Only graphs a) and d) are symmetric about x = 0 so we onlyconsider them. The vertices are at (0, 10) for a) and (0, 1) for d). Only graph d) has thevertex on the same place as the function y = |3x| + 1.

Solution: d.

71


2. Circle the graph of y = −|3x| + 10.

a) b) c) d)

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

One way to find the proper graph is to compute a few points using the formula givenand compare them with corresponding points from the graphs. We see now:f(0) = −|3 × 0| + 10 = 10 - it matches graph a).We compare it with the other graphs (instructive): for b) the graph shows f(0) = 4 - itis not the same value (10), the same way f(0) = −8 for graph c) and f(0) = 1 for d).Therefore, the graph is a).

Solution: a.

3. Circle the graph of y = |3 − x| + 1.

a) b) c) d)

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

We find the vertex and axis of symmetry. We need to find what the axis of symmetryis for this function. The lowest point of the function is when the absolute value has itsminimum. The minimum value for the absolute value function is 0. For this function itcan be obtained when x = 3, i.e. we solve |3 − x| = 0. We find that x = 3. This is our axisof symmetry.Looking at the picture we see the graphs a) and c) are symmetric about x = 0. Graph b) isnot symmetric so it is not a solution. We are left with choice d). This graph is symmetricabout x = 3.

Solution: d.

4. Circle the graph of y = |2x + 4| − 3.

a) b) c) d)

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

We find the minimum value the function can have: it is when the absolute value has itsminimum 0. That minimum is when |2x + 4| = 0 or x = −1/2. Then, f(−1/2) = −3. Theonly graph with minimum value of f(x) = −3 is b).

72


Note we can use any other point to find the graph. For example,we compute the valueof all functions at x = −2 and compare them with the analytical value which is y(−2) =|2× (−2)+4|− 3 = −3. Graph a) shows value of (+3), graph b) - (−3) and graph c) - (+6).Graph d) is undefined for the point x = (−2). The only graph which matches is b).

Solution: b.

5. Circle the graph of y = |3(x − 5)| − 7.

a) b) c) d)

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

–10

–8

–6

–4

–20

2

4

6

8

10

–10 –8 –6 –4 –2 2 4 6 8 10x

We need to consider graphs b), c) and d).We again will use the minimum value of thefunction. It is (−7). The only function with minimum value of (−7) is c) so it is thesolution.

Solution: c.

73


3.8.3 BBc3: Solve equations of the type |ax + b| + c = d.

Solve each equation for x if possible. If it is not possible, state why.

1. For solving |2x + 4| + 3 = 9 we first isolate the absolute value on one side of theequality:

|2x + 4| = 6

We use the fact that absolute value is the number without its sign. Therefore, weneed to solve two equations:2x + 4 = 6and2x + 4 = −6(we use the fact | + 6| = | − 6| = 6). The solutions are:

2x = 6 − 4,

x1 = 1.

2x = −6 − 4,

x2 = −5.

The solutions are x1 = 1 and x2 = −5.

2. |5x + 3| − 3 = 8.

|5x + 3| = 8 + 3 = 11.

The equations are: 5x + 3 = 11 and 5x + 3 = −11.

5x + 3 = 11,

5x = 8,

x1 =8

5.

5x + 3 = −11,

5x = −14.

x = −14

5.

The solutions are x1 = 8/5 and x2 = −14/5.

74


3. |x − 4| + 3 = 1.We isolate the absolute value expression:|x − 4| = 1 − 3 = −2.We have on the left side a positive number, and on the right side — negative. Thesetwo numbers could never be equal. Therefore, this equation does not have a solu-tion.

4. |3x − 4| + 7 = 11.|3x − 4| = 11 − 4.

The equations are 3x − 4 = 4 and 3x − 4 = −4.

3x − 4 = 4

3x = 8,

x1 =8

3.

3x − 4 = −4,

3x = 0,

x2 = 0.

The solutions are x1 = 8/3 and x2 = 0.

5. |4 − 3x| − 3 = 8.|4 − 3x| = 8 + 3 = 11.

The equations are 4 − 3x = 11 and 4 − 3x = −11.

4 − 3x = 11,

3x = −7,

x1 = −7

3.

4 − 3x = −11,

3x = 15,

x2 = 3.

The solutions are x1 = −7/3 and x2 = 3.

75

3.9. LINEAR INEQUALITIES CHAPTER 3. ALGEBRA

3.9 Linear Inequalities

3.9.1 BBd1: Solve or graph linear inequalities in one variable.

Solve each inequality algebraically and graphically. Give your answer algebraically andgraph it on a number line.

Example: If 2x + 5 ≤ 9 then x ≤ 2.

1. 3x + 4 < 7.On the first step we add (−4) to both sides of the inequality:

3x + 4 + (−4) < 7 + (−4) = 3.

On the second step we multiply with the reciprocal of coefficient in front of x:

3x × 1

3< 3 × 1

3,

x < 1.

x

1

+-

We solve all the inequalities the same way. To make it easier, we always transfer x tothe side it will have a positive coefficient. When we multiply by its reciprocal, thisnumber is positive, it does not change the way of the inequality: We have greater orequal, or less or equal ...

2. 3x + 7 ≤ −4x + 8.We add this time (4x − 7) to both sides.

3x + 7 + (4x − 7) ≤ −4x + 8 + (4x − 7),

7x × 1

7≤ 1 × 1

7,

x ≤ 1

7.

76

CHAPTER 3. ALGEBRA 3.9. LINEAR INEQUALITIES

x

xxxx

+-

1/7xxxxxxxxxxxxxx

3. 9 − x ≥ 5x + 3.

9 − x + (x − 3) ≥ 5x + 3 + (x − 3),

6 × 1

6≥ 6x × 1

6,

1 ≥ x.

x

xxxx

+-

xxxxxxxxxxxxxx

1

4. 12x + 4 < 4x − 8.

12x + 4 + (−4x − 4) < 4x − 8 + (−4x − 4),

8x × 1

8< −12 × 1

8,

x < −3

2.

-3/2

X +-

5. 11x + 3 > 19x − 5.

11x + 3 + (−11x + 5) > 19x − 5 + (−11x + 5),

8 × 1

8> 8x × 1

8,

1 > x.

77


x

1

+-

78


BBd2: Solve or graph linear inequalities in two variables.

Solve each inequality for the indicated variable and graph your solution.

When solving an equation or inequality in one variable, we have zero, one or morenumbers as solutions. We can look at these numbers as points on the number line and wecan plot them. In this case, we can think of them as a set of solutions.Analogously, we can think of the solutions of inequality in two variables as the set con-taining all ordered pairs of numbers which make the inequality true sentence. When plot-ting the solution, we have it as a region in (x, y) plane. When solving linear inequality intwo variables, the solution set contains half of plane.

1. Solve for y and graph your solution: 2x − 3y < 4x − 9.

2x − 3y + (3y − 4x + 9) < 4x − 9 + (3y − 4x + 9),

(−2x + 9) × 1

3< 3y × 1

3,

−2

3x + 3 < y,

y +2

3x − 3 > 0.

We see we have the line y + 2/3x − 3 = 0 as a border between two regions in x, yplane — y +2/3−3 > 0 and y +2/3x−3 < 0. To find which one is bigger, we chooseany point which is not on the line. One good test point is (0, 0), if it is not on theline. We check:

0 +2

3× 0 − 3?0.

The answer is that this point is in the half plane y +2/3x−3 < 0. The solution of theinequality is the other half of the plane. The line is dashed because we have strictinequality and therefore, the border (the line) is not part of the solution.

THE TOP HALF OF PLANE IS THE SOLUTION.

X

y

Y = - 2/3X + 3

79


2. Solve for x and graph your solution: 5x − 3y ≤ 4x − 9.

5x − 3y − 4x + 9 ≤ 4x − 9 − 4x + 9,

x − 3y + 9 ≤ 0,

and solved for xx ≤ 3y − 9.

We check x ≤ 3y − 9, using again (0, 0) and find:

0 ≤ 3 × 0 − 9 FALSE!

therefore, the half of plane containing (0, 0) is not the solution of the problem. Thedividing line is part of the solution because we have a less or equal sign.

y=1/3x-3

THE SOLUTION IS THE BOTTOMPART OF THE PLANE (UNDER THE LINE)

X

Y

3. Solve for y and graph your solution: 2x − 3y > x − y + 99.

2x − 3y + (−2x + y) > x − y + 99 + (−2x + y),

−2y > −x + 99,

y < 1/2x − 99

2.

We check with point (0, 0):

0?1

2× 0 − 99

2,

0 > −99

2.

We see that ordered pair (0, 0) is not a solution of the inequality, therefore that partof the plane is not the solution set. The border line is dashed because it is not partof the solution.

80


BOTTOM HALF OF THE PLANE IS THE SOLUTION.

x

y

y=1/2x-99/2

4. Solve for x and graph your solution: 2x + 13y < 4x − 9.

2x + 13y + (−2x + 9) < 4x − 9 + (−2x + 9),

(13y + 9) × 1

2< 2x × 1

2,

13

2y +

9

2< x.

We use the ordered pair (x, y) = (0, 0) and check if it makes the above inequality atrue sentence. It is false because

13/2 × 0 + 9/2 > 0 = x

but we need the inequality the opposite way.Note here and in #5 we have x as a function of y so that the axes changed theirplaces.

x=13/2

y+9/2

THE SOLUTION IS TOP PART OF THE PLANE.x

81


5. Solve for x and graph your solution: 2x − 3y ≥ 4x − 9.

2x − 3y + (−2x + 9) ≥ 4x − 9 + (−2x + 9),

−3y + 9 ≥ 4x − 9 − 2x + 9 = 2x,

(−3y + 9) × 1

2≥ 2x × 1

2,

−3

2y +

9

2≥ x.

As usual, we check using a pair of numbers (x, y) = (0, 1):

−3

2× 1 +

9

2?0,

3

2+

9

2= 3 ≥ 0.

This pair makes the inequality a true sentence, so this is the solution half of plane.

x= - 3/2y +9/2

The solution is bottom half of the plane

82


3.9.2 BBe1: Solve consistent 2 x 2 systems by substitution or elimina-

tion.

Solve a pair of equations algebraically. Check your answers.

1. {4x + 5y = 11, 6x + 11y = −5}We want to eliminate variable x. For this, we need to multiply the first equation by3, and the second by 2:Note I use vertical line ”|” to denote that the operation given on right is performedon both sides of the equation — here we multiply, for example, the first equationwith 3 — all coefficients are multiplied with this number. This notation will be usedin many other plases.

4x + 5y = 11 | × (+3),

6x + 11y = −5 | × (−2).

+12x + 15y = 33,

−12x − 22y = 10.

We add these equations and then solve for y:

+12x − 12x + 15y − 22y = 33 + 10,

−7y = 43 | ×(

−1

7

)

,

y = −43

7.

We use the same way to find x, only this time we multiply the first equation with 11and the second with (−5):

+44x + 55y = 121,

−30x − 55y = 25.

14x = 146,

83


x =146

14=

73

7.

We now check our answer:

4 × 73

7+ 5 ×

(

−43

7

)

=292 − 215

7=

77

7= 11,

6 × 73

7+ 11 ×

(

−43

7

)

=438 − 473

7= −35

7= −5.

2. {2x + 5y = 0, 6x + 11y = 5}We will apply for this example the substitution method. We use the first equation toexpress x :

2x = −5y,

x = −5

2y.

Now, we substitute that value in the second equation and solve for y:

6x + 11y = 6 ×(

−5

2y

)

+ 11y = 5,

−15y + 11y = 5,

−4y = 5,

y = −5

4.

On the next step, we use the expression for x (found before) and calculate its value:

x = −5

2y = −5

2×

(

−5

4

)

=25

8

We check our work:

2 × 25

8+ 5 ×

(

−5

4

)

=25 − 25

4= 0,

6 × 25

8+ 11 ×

(

−5

4

)

=75 − 55

4= 5.

3. {x + 5y = 8, x + 11y = −5}We see both equations contain the variable x with coefficient of one, so this system issuitable for solving using the substitution method. We express x in the first equationin terms of y. On the next step, we substitute this value into the second equation andsolve it for remaining variable, in this case, y.

x = 8 − 5y,

84


x + 11y = (8 − 5y) + 11y = −5.

We add (−8) to each side.

6y + 8 = −5 | + (−8),

6y = −13,

y = −13

6.

We now can return to the first equation x = 8 − 5y and compute x:

x = 8 − 5y = 8 − 5 ×(

−13

6

)

,

x =48 − (−65)

6=

113

6.

We now check the solution:

x + 5y =113

6+

[

5 ×(

−13

6

)]

=113 − 65

6=

48

6= 8.

x + 11y =113

6+

[

11 ×(

−13

6

)]

=113 − 143

6=

−30

6= −5.

4. {4x − 5y = −2, 6x + 10y = −5}We will mix the methods in this example. We will use the elimination method onthe first step to eliminate y. For this we multiply first equation by 2 and add it tosecond equation.

4x − 5y = −2 | × 2,

8x − 10y = −4,

6x + 10y = −5,

14x = −9,

x = − 9

14.

Now, we will substitute the value of x we found in one of the equations and solve itfor the second variable, y:

4x − 5y = −2 | + (−4x),

−5y = −2 − 4x |× − 1

5,

y = −1

5(−2 − 4x).

Note that −1/5 = (−1)(1/5) and −2 − 4x = (−1)(2 + 4x), therefore we can get ridof these minus ones by multiplying them together. The process for computing the

85


value of y is shown below.

y =1

5(2 + 4x),

y =1

5

[

2 + 4 ×(

− 9

14

)]

,

y =1

5× 2 × 14 − 4 × 9

14,

y =1 ×−18

5 × 14,

y = − 4

35.

We check our answer:

4x − 5y =[

4 ×(

− 914

)]

−[

5 ×(

− 445

)]

= −2,

6x + 10y = 6 ×(

− 914

)

+ 10 ×(

− 435

)

= −5.

5. {3x + 5y = 1, 6x + 11y = −5}We multiply the first equation by (−2) and add it to the second one. We substitutethe value of y in the first equation and solve it for x:

3x + 5y = 1| × (−2),

−6x − 10y = −2,

6x + 11y = −5,

y = −7,

3x = 1 − 5y,

3x = 1 − 5 × (−7),

3x = 36,

x = 12.

We check our work:

3x + 5y = 3 × 12 + 5 × (−7) = 1,

6x + 11y = 6 × 12 + 11 × (−7) = −5.

It may seem easier to solve a system using the substitution method and comparing itwith the elimination method but it is not. For example, what if we have a system with4 equations and decide to use this method? On the first step, we express one variablein terms of other three. Then, we substitute the result in the other three equations andsimplify. We need to repeat the process again —it means the method is not used for largesystems.

86


3.9.3 BBe2: Interpret 2 x 2 systems graphically .

Graph each of the following systems. Put the coordinates of the point of intersection onyour graph.

Example: {4x + 5y = 1, 6x + 11y = 5}.Solution: x = −1 and y = 1, so (−1, 1) is the point of intersection. Graph each of thefollowing systems. Put the coordinates of the point of intersection on your graph.

Explanation ( applied to all of the exercises below):Recall that solution set for an equation in two variables can be drawn in a plane. For alinear equation, it is a line in the plane. If we have two lines, they (a) can be parallel, (b)can have an intersection point or (c) — can be one and the same line. Then, the solutionset of a system in two variables and two equations is the set of all points, which makeeach of the equations true sentence - i.e. all points in the plane which belong to both lines.Graphically this is the intersection point of the two lines (see the graph below).

4x+5y=1

6x+11y=5

Intersection point (-1,1)

(-1,1)

87


1. {4x + 5y = −11, 6x + 11y = 5}.

4x+5y=-11

6x+11y=5

Intersection point about (-10, 7)

(-73/7, 43/7)

Graphing helps us to find out whether the solution exists. The precision, however,is not high. The analytic method shows that x = −73/7 and y = 43/7.We check:

4 ×(

−73

7

)

+ 5 × 43

7= −292

7+

215

7= −77

7= −11,

6 ×(

−73

7

)

+ 11 × 43

7= −438

7+

473

7=

35

7= 5.

88


2. {2x + 5y = 0, 6x + 11y = −5}

2x + 5y = 0

6x + 11y = -5

(-25/8, 5/4)

We check:

2 ×(

−25

8

)

+ 5 × 5

4= −25

4+

25

4= 0,

6 ×(

−25

8

)

+ 11 × 5

4= −75

4+

55

4= −4.

3. {x + 5y = −8, x + 11y = 5}

x + 5y = -8

x + 11y = 5

(-113/6, 13/6)

We check:

−113

6+ 5 ×

(

13

6

)

= −113

6+

65

6= −48

6= −8,

−× 113

6+ 11 ×

(

13

6

)

= −113

6+

143

6=

30

6= 5.

4. {4x − 5y = 2, 6x + 10y = 5}

89


4x - 5

y = 2

6x +10y = 5

(9/14, 4/35)

We check:

4 × 9

14− 5 × 435 =

18

7− 4

7= 2,

6 × 9

14+ 10 × 435 =

27

7+

8

7= 5.

5. {3x + 5y = −1, 6x + 11y = 5}

3x + 5y = -1

6x +11y = 5

(-12 , 7)

We check:3 × (−12) + 5 × 7 = −36 + 35 = −1,

6 × (−12) + 11 × 7 = −72 + 77 = 5.

90


3.9.4 BBe3: Recognize inconsistent or dependent systems algebraically

or graphically.

Indicate whether each system represents either a pair of parallel lines, the same line, or apair of lines that intersect.

Example: Indicate whether each system represents either a pair of parallel lines, thesame line, or a pair of lines that intersect.{4x + 5y = 1, 8x + 10y = 5}.

Solution of the solved example:We rearrange both equations in slope-intercept form, i.e. solve them for y:

4x + 5y = −11,

5y = −4x − 11,

y = −4

5x − 11

5.

8x + 10y = 5,

10y = −8x + 5,

y = − 8

10+

5

10,

y = −4

5x +

1

2.

We look at the expressions for y. What we see is both have slope −4/5 and differenty-intercepts. This means the lines are parallel.

1. {4x + 5y = −11, 6x + 11y = 5}.We can solve the problem graphically. We plot the two lines and observe whetherthey are parallel, one and the same line or intersect each other. The graph belowshows that the lines intersect each other.

4x + 5y = 116x + 11y = 5

Rearranged equations show the same:

y = −4

5x +

11

5,

y = − 6

11x +

5

11.

91


Slopes are −4/5 and −6/11.For the remaining systems I will show only the rearranged equations and the con-cusion.

2. {x + 5y = 10, 2x + 10y = 20}.

x = −5y + 10,

x = −5y + 10.

This system is one and the same line.

3. {−x + 11y = −8, x − 11y = 5}.

y =1

11x − 8

11,

y =1

11x +

5

11.

This is a system of parallel lines.

4. {4x = 2 + 2y, 8x − 4y = 5}.

x =1

2y +

1

2,

x =1

2y +

5

2.

This is a system of parallel lines.

5. {3x + 5y = −1, 6x + 11y = 5}.

y = −3

5x − 1

5,

y = − 6

11x +

5

11.

This is a system of intersecting lines.

92

CHAPTER 3. ALGEBRA 3.10. VERBAL PROBLEMS

3.10 Verbal Problems

3.10.1 BBf1: Work/rate.

Solve each of the following problems involving rates and work.Example: Two bricklayers are to build a wall. The first can lay 50 bricks per hour and

the second can lay 70 bricks per hour. How long will it take them to construct the wall ifit contains 1200 bricks?Solution:

Suppose only the first builder works on the wall. He will lay 50 bricks in one hour. If hegets a helper, i.e. the second builder comes to work, both will lay 120 bricks in one hour.For two hours they lay 240 bricks, for 3— 360 and so on. The builders need to lay 1200bricks. To find out the time, we divide the total number of bricks to the number of brickswhich the two builders can lay per hour. This way we find:

TIME =Total number of bricks

Number bricks laid per hour

t =1200

120= 10

The two builders need 10 hours to construct a wall containing 1200 bricks.

Definition:Rate is the amount of work done in one unit time by one or more people.

1. It takes one bricklayer 20 hours to lay 1200 bricks, and it takes another bricklayer 16hours to lay 1200 bricks. Working together, how long will it take them to lay 2400bricks?

We look at the formula above. We know the total amount of work to be done but wedo not know the rate at which it could be done. Therefore, we need to find the rateof the two builders together. If the first builder can lay 1200 bricks in 20 hours, thisrate can be found by dividing the number of bricks built by the time for doing so:

RATE =Total number of bricks

Time for building.

We denote the first builder rate by RATE1 and the second builder rate by RATE2.

RATE1 =1200

20= 60 bricks per hour.

93

3.10. VERBAL PROBLEMS CHAPTER 3. ALGEBRA

The first builder rate is 60 bricks per hour. We find the second builder’s rate thesame way:

RATE2 =1200

16= 75 bricks per hour.

The rate at which the two builders work is the sum of the rates of the two builders:

RATE = RATE1 + RATE2 = 60 + 75 = 135 bricks per hour.

The two builders working together have a rate of 135 bricks per hour. They need tolay 2400 bricks. We use now the formula from above:

TIME =Total number of bricks

RATE,

t =2400

135= 17

2

3.

These two builders need 17 hours and 40 minutes to lay 2400 bricks.

2. My hot water faucet fills my bathtub in 20 minutes and my cold water faucet fillsmy bathtub in 16 minutes. If I run both at the same time, how long will it take to fillthe bathtub?

We take another look at the rate definition. We do not know the volume of the bath.Let us consider it 1 unit(i.e. it could be 1 ft3, 1 m3 or we can make another unit forthis particular problem - it is 1 bath). Then, the rate for filling the bath is:

RATE =1 bath

Time to fill one bath

We will denote RATE1 the rate hot water faucet fills the bath and RATE2 the ratecold water faucet fills the bath. Then:

RATE1 =1

20bath per minute,

RATE2 =1

16bath per minute.

If we open both faucets the rate is:

RATE = RATE1 + RATE2 =1

20+

1

16=

9

80bath per minute.

We can now use the formula for computing time:

TIME =1 bath

RATE=

1

9/80.

If I open both hot and cold water faucets, I can fill the bath for 889

minutes, approxi-mately 8 minutes and 53 seconds.

94


3. One secretary can type 120 words per minute, while another can type 130 wordsper minute. Working together, how long would it take them to type a documentcontaining 100,000 words?

The two secretaries working together type 250 words per minute. The time for

typing 100,000 words could be found by the same formula:

TIME =Number of words

RATE=

100, 000

250= 400.

The two secretaries working together can type in 100,000 words for 400 minutes or6 hours and 40 minutes.

4. It takes one groundskeeper an hour to mow a football field, and it takes another anhour and twenty minutes to mow the same field. Working together, how long doesit take to mow the field?

This is an another variation for the above problem. We have one field. The rates formowing the field are ( in minutes):

RATE1 =1

60fields per minute,

RATE2 =1

60 + 20fields per minute.

The rate for the two grounkeepers is

RATE = RATE1 + RATE2 =1

60+

1

80=

7

240fields per minute.

The time for mowing is:

TIME =17

240

= 342

7

The two groundkeepers can mow the field for 3427

minutes, approximately 34 minutesand 17 seconds (if nobody stops to get a sip of water).

5. My neighbor and I want to fill a swimming pool. His hose delivers 15 gallons perminute, and mine delivers 12 gallons per minute. Alas, the pool leaks 3 gallons perminute. When full, the pool contains 10,000 gallons of water. How long will it takeus to fill the pool?

Let’s see what happens after one minute if I and my neighbor start filling the swim-ming pool. We will have in the pool 12 gallons of water coming from my hose, 15

95


gallons of water coming from my neighbors hose, but there will be 3 gallons less be-cause of a leek. Effectively, there will be 24 gallons in the pool. If we denote RATE1

— the rate of my hose, RATE2 — rate of my neighbor’s hose and RATE3 the rate atwhich the pool leaks, we can write:

RATE = RATE1 + RATE2 − RATE3 = 24 gallons per minute.

Now, we can compute the time for filling the swimming pool:

TIME =10000

24= 416

2

3

We need 41623

minutes to fill the swimming pool or 416 minutes and 40 seconds.

96


3.10.2 BBf2: Distance/rate/time.

Solve each of the following problems involving distance, rate and time.Example: If I can run 7.5 miles per hour for two hours, how long will it take me to run

10 miles?We make use of the formula connecting distance, speed and time:

distance = speed × time.

In this example the distance is 10 miles, the speed is 7.5 miles per hour, and we do notknow the time. We use the expression for time :

time =distance

speed=

10

7.5= 1

1

3

The time for running 10 miles is 113

hours.

1. A car travels 10 miles in 8 minutes. What is its average speed in miles per hour?

There are two ways to solve this problem. We can compute what is the averagespeed per minute and then, knowing that one hour has 60 minutes, we multiplythis number by 60, and thus we find the average speed per hour. Here are thecomputations:Rate per minute (denoted by RPmin):

RPmin =10

8=

5

4miles per minute.

We compute the speed:

speed = 60 × RPmin = 60 × 5

4= 75 mph.

The speed is 75 mph.

The other way is to express 8 minutes in hours:

8 minutes =8

60hours =

2

15hours.

We can now compute the average speed using the main formula directly:

speed =distance

time=

10

2/15

speed = 75 mph.

NOTE: We always need to use one and the same units for all different variables inthe formula, i.e. if one variable is in miles, the other need to be in miles per unittime, the units time must be the same and so on.

97


2. A car travels 10 miles at an average speed of 45 miles per hour. How many minutesdoes this take?

We make use of the same formula (speed×time=distance):

time =distance

speed=

10

45=

2

9ho.

Because we need the time in minutes we multiply the found time in hours by 60,the number of minutes in 1 hour:

time (min) = time (ho) × 60 =2

9× 60 =

40

3= 13

1

3min.

Above, we denote [time(min)] the time for traveling 10 mi in minutes and [time(ho)]the same in hours.It takes 131

3minutes for a car with average speed 45 mph to travel 10 miles.

3. It takes 11 minutes for a leaf to drift 1/4 mile down a river. How fast is the riverflowing in feet per second?

We will express the time in seconds and the distance in feet, then the answer we findwill be in feet per second. There are 5280 feet in a mile.

speed × time = distance,

speed =1/4 × 5280

11 × 60

=1320

660= 20 feet/sec.

The speed of the current is 20 feet per second.

4. A river is flowing at 2 feet per second. How long will it take for a boat to drift 100feet? Since all data is in desired units, we write:

time =distance

speed

=100

2= 50 sec.

The boat drifts 100 feed for 50 seconds.

98


5. Jackie Joyner-Kersee once ran 800 meters in 2 minutes and 17.61 seconds. What washer average speed in kilometers per hour?

We express 800 meters in kilometers 800 m= 0.8 km. We express 2 min 17.61 sec inhours: 2 min 17.61 = (2× 60 + 17.61)/3600 ho. We now compute the average speed:

speed =distance

time

=0.8

137.61/3600= 20.929 km/ho.

Jackie Joyner-Kersee speed was 20.929 km/ho.

99


3.10.3 BBf3: Mixture.

Solve each of the following problems involving rates and work.Example: I have one gallon of a 12% alcohol solution and one gallon of a 20% alcoholsolution. How much of each should I mix together to get one gallon of a 15% alcoholsolution?

Example solution:Recall when we have two unknown quantities, we need two different equations to findthe amounts. First, we need to have one gallon of the solution. If we denote with x thequantity of 12% solution and with y the quantity of 20% solution, we know when we addthem together, their sum is one gallon solution. We translate it:

x + y = 1.

We need 15% solution. If we have 1 gallon of 12% solution, then the amount of alcoholit contains is 12% × 1 gallons, but we must take another quantity — x gallons. In thisquantity, we have 12% × x gallons of alcohol. The same way we determine in y gallons20% solution we have 20% × y gallons of alcohol. We know when we add these together,we need to have 1 gallon 15% solution. The alcohol in it is 15%× 1 gallons. The equatioinis

.12x + .2y = .15.

Solving above equations together, we find x = 0.625 and y = 0.375.We need 0.625 gallons 12% solution and 0.375 gallons 20% solution.Remember, the check of the answer is part of the solution:

0.625 + 0.375 = 1.000 liters,

12% × 0.625 + 20% × 0.375

= 0.075 + 0.075 = 0.150 = 15% × 1liters.

1. My radiator contains 4 gallons of a 35% antifreeze solution. How much should Ireplace with pure antifreeze to get 4 gallons of a 50% antifreeze solution?

We start with 4 gallons of 35% solution. We take out x gallons of it and put back xgallons of 100% solution. The result is 4 gallons 50% solution. From the picture, wecan visualize the equation:

(4 × 0.35) − (x × 0.35) + x = 4 × 0.5.

100


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4 gal 35 %solution

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_ x gal 35 %solution

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+x gal 100 %solution

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=

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4 gal 50%solution

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

We replace xgal 35 % solution

xxxxxxxxxxxxxxxxxxxxxxx

xxxx

xxxx

with x gall100 %

xxxxxxx

We solve the resulting equation and find x = 1213

gallons.

2. Ten gallons of a 5% alchohol solution is mixed with fifteen gallons of a 23% alcoholsolution. What is the concentration of alcohol in the resulting solution?

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10 gall 5 %solution

15 gall 23 %solution

+

25 gall x % solution=

From the graph we see that total amount of solution is 25 gallons. We see also thealcohol in it is (5% × 10) + (23% × 15). Recall the concentration is computed bydividing the amount of alcohol by the amount of the solution:

concentration =amount of alcohol

amount solution× 100,

101


concentration =0.05 × 10 + 0.23 × 15

10 + 15× 100 = 15.8%.

3. Peanuts cost $1.50 per pound, and walnuts cost $3.25 a pound. I want to blendthem to get 10 pounds of nuts that sells for $2.50 per pound. How many pounds ofpeanuts and how many pounds of walnuts should I use?We denote with x the amount of peanuts and y the amount of walnuts. The infor-mation is organized in the chart below.

Peanuts Walnuts MixQuantity x y 10Price $1.50 $3.25 $2.50Cost $1.50x $3.25y $2.50 × 10

We can translate the chart into equations: the quantities must add to 10 (x + y = 10)and the costs must match (1.5x+3.25y = 10×2.5). We find then we need to use 30/7pounds of peanuts and 40/7 pounds of walnuts.

4. Cinnamon sugar is made by blending sugar, at $0.40 per pound with cinnamon at$40.00 per pound. Cinnamon sugar sells for $10.00 per pound. How much cinna-mon is in each pound?

We can use the same kind of chart to solve this problem. The information is givenbelow.

Cinnamon Bl. sugar Cinn. shugarQuantity x y 1Price $40.00 $0.40 $10Cost $40.00x $0.40y $10 × 1

We write down the equations as in #3:

x + y = 1, total amount40x + 0.4y = 10 × 1 cost equation.

After solving the equations, we find we need 8/11 pounds of blended sugar and3/11 pounds of cinnamon.

5. Fuel for a two cycle engine is made by blending gasoline with oil. The gasoline costs$2.20 per gallon and the oil costs $4.00 per gallon. How much should one gallon ofthe mixture cost if it contains 32 times as much gasoline as oil?We know here the ratio in which we blend the gasoline and oil. If we use 32 gallonsof gasoline and one gallon of oil, we will have 33 gallons of mixture. The cost ofprice of this mixture is 32 × $2.20 + $4.00 = $74.4. We divide this dollar amount bythe number of gallons and find the cost per gallon: $2.26.

102


3.10.4 BBf4: Number (reciprocal, digit, consecutive, integer, etc.).

Solve each of the following problems involving numbers.Example: The sum of three consecutive integers is 24. What are these integers?

Example solution:Let denote the first integer with x, then the next consecutive integer is (x+1) and the nextis (x + 2). We know the sum of the three of them is 24. The equation is:

x + (x + 1) + (x + 2) = 24.

We solve the equation for x and find x = 7.Answer: 7, 8 and 9.Check:

7 + 8 + 9 = 24.

1. 3 more than twice a number is 17. What is the number?

We denote the number with x. We translate into mathematical language: three morethan twice is (3 + 2x) and the answer is 17:

3 + 2x = 17.

After solving equation we find that the number is 7.Answer: 7.Check:

3 + 2 × 7 = 17.

2. A real number larger than 1 is added to its reciprocal and the result is 29/10. Whatis this number?

We denote the number with x. We know the sum of the number (x) and its reciprocal(1/x) is 29/10. The equation is

x +1

x=

29

10.

We multiply this equation with x and the resulting equation is quadratic:

x2 − 29

10x + 1 = 0,

10x2 − 29x + 10 = 0.

We solve it and find x1 = 5/2 and x2 = 2/5. These two numbers are reciprocals soonly one is bigger than one.Answer is x = 5/2.Check:

5

2+

1

5/2=

5

2+

2

5

=25

10+

4

10=

29

10.

103


3. The sum of three consecutive odd integers is 375. What are these integers?

We denote the smallest integer with x, therefore next integer is (x + 2) and the thirdone is (x + 4). We have the equation:

x + (x + 2) + (x + 4) = 375.

We solve the equation for x and find x = 123.The integers are 123, 125 and 127.Check:

123 + 125 + 127 = 375.

4. The product of two integers is 21 and their sum is 10. What are these integers?

We denote one of the numbers with x and the other with y. We know their sum is10 so one equation is x + y = 10. The product is 21 so the equation is xy = 21. Thesystem is:

x + y = 10,

xy = 21.

The way to solve this nonlinear system is by expressing y in the first equation:

y = 10 − x.

We substitute y in the second equation and solve.

x(10 − x) = 21,

10x − x2 − 21 = 0,

x2 − 10x + 21 = 0.

We find the solutions of this equation are 3 and 7. They are the numbers with a sumof 10 and product of 21. Thus, the system of equations has two solutions but thereis only one answer to the question since 3 + 7 = 7 + 3 and 3 × 7 = 7 × 3.Answer: 3 and 7. The explanation above cleans up the check of the answer.

5. The product of two real numbers is 3 and their sum is 5. What is the larger of thesetwo real numbers?

We use the same method as in #4. I will write down the system:

x + y = 5,

xy = 3.

We express y from the first equation:

y = 5 − x.

104


Using the second equation we write:

3 = xy = x(5 − x) = 5x − x2,

x2 − 5x + 3 = 0.

This equation has irrational solutions. The discriminant is D = (−5)2 − 4 × 1 × 3 =25 − 12 = 13. The larger number is

x =5 +

√13

2.

The other number is

y =5 −

√13

2.

We check:

x + y =5 +

√13

2+

5 +√

13

2= 5,

xy =5 +

√13

2× 5 −

√13

2=

52 − (√

13)2

4=

25 − 13

4= 3.

6. Twice one real number plus six times another is 13 while the average of these num-bers is 4. What are these numbers?

We denote first number with x and second with y. We use the information now,translatingit into equations: Twice the first number plus 6 times second is 13: 2x+6y = 13. Theaverage is 4: x+y

2= 4. The system:

2x + 6y = 13,x + y

2= 4.

We multiply the second equation by 4:

2x + 2y = 16.

We subtract that equation from the first one:

2x + 6y − (2x + 2y) = 13 − 16 = −3.

We solve for y:4y = −3,

y = −3

4.

The corresponding x is

x =35

4.

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We check:x + y

2=

35/4 + (−3/4)

2= 4,

2x + 6y = 2 × 35

4+ 6 ×

(

−3

4

)

=35

2− 9

2= 13.

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3.10.5 BBf5: Area and perimeter.

Solve each of the following problems involving numbers.Example: The area of a circle is 25π square feet. What is the circumference of this

circle?

Example solution:Recall that the area A, of a circle is

A = πr2,

where r is the radius and the circumference, l, is

l = 2πr.

We find the radius r from the expression for the area: r = 5. We use now the secondformula and find that the circumference is 10π. Answer: 10π feet.

1. The circumference of a circle is 15 feet. What is the area of this circle?

We use the same formulas but in opposite way. From the circumference formula wefind the radius of the circle.

l = 2πr,

r =l

2π=

15

2π.

Using the formula for the area we find:

A = πr2 = 3.14 × 2.392 = 17.95

The area is 17.92 square feet.

2. The perimeter of a square is 64 meters. What is the area of this square?

Recall the formulas for the perimeter p = 4 × s and the area is A = s2, where s isthe length of a side of the square. We use the formula for the perimeter and find theside of the square:

p = 4 × s,

s =p

4=

64

4= 16.

Using the formula for the area we compute

A = s2 = 162 = 256.

The are of a square with perimeter 64 meters is 256 square meters.

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3. The area of an equilateral triangle is 16 square inches. What is the perimeter of thistriangle?

Recall the formula of the area of a triangle is A = 12aha. In this formula a is one side

of the triangle, and ha is the perpendicular to the same side. In equilateral triangleis the same. We only need to find the length of the perpendicular. To do so, we usethe right angle triangle in which the legs are AH and HC, the hypotenuse is AC.Weknow that AC = a and AH = a/2. We do not know the length of the third side. Weuse the Pythagorean theorem and write:

AH2 + HC2 = AC2,(a

2

)2

+ h2a = a2,

h2a = a2 −

(a

2

)2

=3a2

4,

ha =

√3a

2.

We can now use the formula of the area and the height we found.

A =1

2a × ha =

1

2a ×

√3a

2,

A =

√3

4a2.

We find a:

a =

√

4A√3

=

√

4 × 16√3

≈ 6.08 inches.

The perimeter is the sum of the lengths of the sides, in this case — three times theside:

p = 3 × a = 3 × 6.08 ≈ 18.24 inches.

The perimeter is 18.24 inches.

4. A rectangle is twice as wide as it is high, and the perimeter is 12 meters. What is thearea of this rectangle?

Let denote the high with x. Then, the wide is twice the high, i.e. (2x). The perimeteris the double sum of the two : p = 2(x + 2x). It is 12. We write:

6x = 12

x = 2

The high is 2 meters and the wide is 4 meters long. The area of the rectangle is theproduct of the wide and high, and is 8 square meters.The area of the rectangle is 8 square meters.

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5. The area of a rectangle is six square inches and the perimeter is 10 inches. What arethe dimensions of this rectangle.

We denote the wide of the rectangle with x and the high with y. We know that theperimeter is doubled sum of the sides and is 10: 2(x + y) = 10. The area is theproduct of the sides and is 6: xy = 6. The system is

2(x + y) = 10

xy = 6

After solving the system we find x = 2 and y = 3.The dimensions of the rectangle are 2 inches and 3 inches.

6. If we quadruple the area of a circle, by what factor do we increase its circumference?

Let A is the area of the initial circle and it is

A = πr2

Let denote r1 the radius of quadruple circle. We use the same formula:

4A = πr21

We substitute A in this expression and obtain:

4A = 4π(r2) = πr21

We solve the above expression for r1:

r1 = 2r

We denote with l the circumference of the original circle and with l1 of the quadru-pled one. We express now the circumference of that bigger circle in terms of thesmaller one:

l1 = 2πr1 = 2π(2r) = 2(2πr) = 2l

If we quadruple the ares of a circle, its circumference is doubled.

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3.10.6 BBf6: Other miscellaneous problems.

Solve each of the following problems involving distance, rate and time.

1. I own two cars. One gets 22 miles to the gallon of gasoline, and the other gets 32miles to the gallon of gasoline. I drove 720 miles and used 30 gallons of gasoline.How many miles was each car driven?

We denote with x the miles first car was driven and y — the other one respectively.The first car (Car 1) runs 22 miles a gallon and the second one — 32. If the first car isdriven x miles, it will consume x/22 gallons gasoline. For the second one the valueis y/32 gallons. We can organize the information in the chart below:

Car 1 Car 2 TotalsMiles x y 720 miMiles per gallon 22 32 -Gas x/22 y/32 30 gal

There are two relations in this chart — one is the total distance traveled, and theother is total gasoline used. On the first row, the distances traveled are x and y.We add them anf the total is 720 miles. In the third row, we can add the gasolineconsumed by a car and the total is 30 gallons. The system is:

x + y = 720,x

22+

y

32= 30.

After solving this system we find x = 528 and y = 192.The car getting 22 miles a gallon traveled 528 miles and the other one traveled 192miles.

2. Advance tickets to a play cost 7 dollars, and tickets at the door cost 10 dollars. If 300tickets were sold and there were 2700 dollars of income, how many tickets of eachtype were sold?

Let’s denote x the number of the tickets sold in advance and y — at the door. Weorganize all the information in the chart below:

In advance At the door Total# of tickets x y 300Price $7.00 $10.00 —Income $7.00x $10.00y $2700

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Again, we see two relations we can use to find out how many tickets of each kindwere sold — the number of the tickets and the incomes. The system is:

x + y = 300 # of tickets equation,

7x + 10y = 2700 income equation.

After solving the system we find x = 100 and y = 200.There were 100 tickets sold in advance and 200 tickets sold at the door.

3. It takes 6 hours for a plane to fly from west to east from Los Angeles to New YorkCity, and it takes 8 hours for the plane to fly back, owing to the constant west toeast jet stream. If it is 3200 air miles between cities, how fast is the jet stream windblowing?

We make a drawing for better understanding the problem. It is given below:

L. A. N. Y. City

plane jet

plane jet

distance = 3200 miles

time = 6 hours

time = 8 hours

The distance in both directions is 3200 miles. If we denote the speed of plane in stillair by pl and the speed of the jet stream by j, then from LA to New York the speedof the plain is pl + j and the time for flying is 6 hours. In the opposite direction, thespeed is pl − j and the time is 8 hours. We have two relations:

6(pl + j) = 3200,

8(pl − j) = 3200.

We can easily solve this system. The solution is pl = 1400/3 and j = 200/3.The speed of the jet stream is 200/3 mph or 66.67 mph.

4. One child is 5 times as old as another. The sum of their ages is 18 years. How oldwill the younger child be in three years?

Let denote the age of the yonger child now by x. Then, the age of the other child is5x. The sum of the two ages is 18. We have the equation:

5x + x = 6x = 18.

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We solve the equation and find x = 3. After three years this child will be three yearsolder.The age of the child after three years will be 6 years.

5. One leg of a right triangle is 3 times as long as the other. The area of the triangle is6 square meters. What is the perimeter of this right triangle?

Recall the formula for the area A = 1/2× ab of a right triangle, where a and b are thelegs. The shorter leg is a and the longer one is 3a. Using this information, we find a:

1

2a × 3a = A = 6,

1

23a2 = 6.

We find a = 2 or a = −2. The length cannot be negative so we have only onesolution: a = 2. Then, the legs are 2 meters and 6 meters.We need now to find the third side. We use the formula for the sides in the righttriangle: c2 = a2 + b2. In our case, b = 3a, therefore

c2 = a2 + b2

= a2 + (3a)2 = 22 + 62 = 40.

We find c = 2√

10. The length cannot be negative so we do not consider the negativesolution of the equation for the hypothenuse. The perimeter of the right triangle is

p = a + b + c = 2 + 6 + 2√

10

= 8 + 2√

10.

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3.10.7 BBf5: Area and perimeter.

Solve each of the following problems involving numbers.Example: The area of a circle is 25π square feet. What is the circumference of this

circle?

Example solution:Recall that the area A, of a circle is

A = πr2,

where r is the radius and the circumference, l, is

l = 2πr.

We find the radius r from the expression for the area: r = 5. We use now the secondformula and find that the circumference is 10π. Answer: 10π feet.

1. The circumference of a circle is 15 feet. What is the area of this circle?

We use the same formulas but in opposite way. From the circumference formula wefind the radius of the circle.

l = 2πr,

r =l

2π=

15

2π.

Using the formula for the area we find:

A = πr2 = 3.14 × 2.392 = 17.95

The area is 17.92 square feet.

2. The perimeter of a square is 64 meters. What is the area of this square?

Recall the formulas for the perimeter p = 4 × s and the area is A = s2, where s isthe length of a side of the square. We use the formula for the perimeter and find theside of the square:

p = 4 × s,

s =p

4=

64

4= 16.

Using the formula for the area we compute

A = s2 = 162 = 256.

The are of a square with perimeter 64 meters is 256 square meters.

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3. The area of an equilateral triangle is 16 square inches. What is the perimeter of thistriangle?

Recall the formula of the area of a triangle is A = 12aha. In this formula a is one side

of the triangle, and ha is the perpendicular to the same side. In equilateral triangleis the same. We only need to find the length of the perpendicular. To do so, we usethe right angle triangle in which the legs are AH and HC, the hypotenuse is AC.Weknow that AC = a and AH = a/2. We do not know the length of the third side. Weuse the Pythagorean theorem and write:

AH2 + HC2 = AC2,(a

2

)2

+ h2a = a2,

h2a = a2 −

(a

2

)2

=3a2

4,

ha =

√3a

2.

We can now use the formula of the area and the height we found.

A =1

2a × ha =

1

2a ×

√3a

2,

A =

√3

4a2.

We find a:

a =

√

4A√3

=

√

4 × 16√3

≈ 6.08 inches.

The perimeter is the sum of the lengths of the sides, in this case — three times theside:

p = 3 × a = 3 × 6.08 ≈ 18.24 inches.

The perimeter is 18.24 inches.

4. A rectangle is twice as wide as it is high, and the perimeter is 12 meters. What is thearea of this rectangle?

Let denote the high with x. Then, the wide is twice the high, i.e. (2x). The perimeteris the double sum of the two : p = 2(x + 2x). It is 12. We write:

6x = 12

x = 2

The high is 2 meters and the wide is 4 meters long. The area of the rectangle is theproduct of the wide and high, and is 8 square meters.The area of the rectangle is 8 square meters.

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5. The area of a rectangle is six square inches and the perimeter is 10 inches. What arethe dimensions of this rectangle.

We denote the wide of the rectangle with x and the high with y. We know that theperimeter is doubled sum of the sides and is 10: 2(x + y) = 10. The area is theproduct of the sides and is 6: xy = 6. The system is

2(x + y) = 10

xy = 6

After solving the system we find x = 2 and y = 3.The dimensions of the rectangle are 2 inches and 3 inches.

6. If we quadruple the area of a circle, by what factor do we increase its circumference?

Let A is the area of the initial circle and it is

A = πr2

Let denote r1 the radius of quadruple circle. We use the same formula:

4A = πr21

We substitute A in this expression and obtain:

4A = 4π(r2) = πr21

We solve the above expression for r1:

r1 = 2r

We denote with l the circumference of the original circle and with l1 of the quadru-pled one. We express now the circumference of that bigger circle in terms of thesmaller one:

l1 = 2πr1 = 2π(2r) = 2(2πr) = 2l

If we quadruple the ares of a circle, its circumference is doubled.

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