Chapter 7 Review - Precalculus

- SUBSTITUTION- ELIMINATION- PARTIAL FRACTIONS- INEQUALITIES

CHAPTER 7 - REVIEW

Substitution

Example

Solve the system by the method of substitution.

52

4

yx

yx (1)

(2)

Solve (1) for x

4yx

Substitute this result into (2)

52)4( yy

543 y

3y

Back substitute to solve for x

4yx 1

Check solutions

5)3(21

431

)3 ,1(

Example

Solve the system by the method of substitution.

03

022yx

yx (1)

(2)

Solve (1) for x

yx 2


0)2(3 2 yy

0)6( yy

0y


OR 6y

yx 2

yx 2

0

12

Check solutions

000

000

06)12(3

0)6(2122

)6 ,12(

)0 ,0(

Example

-10 -5 0 5 10

-10

-5

0

5

10Solve the system graphically

2

12

yx

yx

Solve each equation for y

2/)1( xy

2xy

(1)

(2)

(1)(1)

(2) (2)

Graph each and find the intersection

(5, 3)

0 2 4 6 8 10 12 14 16 18 20

-10

-5

0

5

10

Example

Solve the system graphically

2

1

2

101142

yx

xy (1)

(2)

Solve each equation for y

)114( xy

2/)1( xy

(1)

(2)

Graph each and find the intersection

(1)

(2)

(15, 7)

(3, 1)

Example

A small fast food restaurant invests $5000 to produce a new fooditem that will sell for $3.49. Each item can be produced for $2.16.How many items must be sold to break even?

x16.2$5000$Cost

Let x be the number of items produced

x49.3$Revenue

Cost - RevenueProfit

)16.25000(49.3)( xxxP

Set profit equal to 0

0)16.25000(49.3 xx

500033.1 x

units 3760x

0 1000 2000 3000 4000 5000 6000

-6000

-5000

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

P(x)

Break even point

units 3760x

Method of Elimination

Example

Solve the system by the method of elimination.

1352

253

yx

yx (1)

(2)

Add equations (1) & (2)

253 yx

1352 yx

15 5 x

3x

Back substitute to solve for y

25)3(3 y

5

7y

Check solutions

135

75)3(2

25

75)3(3

5

7 ,3

Example


1053

127

yx

yx (1)

(2)

Multiple (1) by -3

36213 yx

10 53 yx+

26 26 y

1y


10 )1(53 x

5 x

Check solutions

10)1(5)5(3

12)1(75

)1 ,5(

Example


952

4113

yx

yx (1)

(2)

Multiple (1) by 2 and (2) by 3

8226 yx

27156 yx+

35 7 y

5 y


4 )5(113 x

17 x

Check solutions

9)5(5)17(2

4)5(11)17(3

)5 ,17(

Interpretation of solutions

A linear system of equations is called consistent if it has at least one solution.

A consistent system with exactly one solution is independent. If it has infinitely many solutions it is dependent.

A system is called inconsistent if it has no solutions

Example

Two planes start from LA International airport and fly in opposite directions.The second plane leaves 30 minutes after the first plane, but its speed is

80 kilometers per hour faster than the first plane. Find the airspeed of each Plane if 2 hours after the first plane departs the planes are 32oo km apart.

Let v1 be the speed of the first plane and v2 be the speed of the second one.

km 3200hours) 1.5(hours) 2(distance 21 vv

8012 vv

km 3200)80(5.12 11 vv

30803.5 1 v

km/h 8801 v

8012 vv

km/h 9602 v

Example

A total of $32,000 is invested in two municipal bonds that pay 5.75%and 6.25% simple interest. The investor wants an annual interest Income of $1900 from the investments. What amount should be

invested in the 5.75% bond?

Let x be the amount invested in the 5.75% bond.

What is the amount invested in the 6.25% bond? x000,32

1900)000,32(0625.00575.0 incomeInterest xx

100005.0 x 000,20x

$20,000 should be invested in the 5.75% bond and 12,000 in the 6.25% one

Example

Use any method to solve the following system

2

1637

yx

yx (1)

(2)

Solve (2) for x

2yx


163)2(7 yy

3010 y

3y


2yx 1

Check solutions

231

16)3(3)1(7

)3 ,1(

Example


175

634

yx

yx (1)

(2)

Multiple (1) by 5 and (2) by 4

301520 yx

42820 yx+

26 13 y

2 y


6 )2(34 x

3 x

Check solutions

1)2(7)3(5

6)2(3)3(4

)2 ,3(

Example


52

43

2

2

1

yx

yx (1)

(2)

Multiple (1) by 6

24)2(2)1(3 yx

2323 yx (3)

Solve (2) for x

52 yx

Plug this result into (3)

232)52(3 yy

88 y 1y


52 yx 7 x

(7, 1) satisfies both equations

Example

Solve the system of linear equations

1

232

142

zyx

zyx

zyx (1)

(2)

(3)

Add (1) and (3)

142 zyx1 zyx+

053 yx (4)

Multiply (1) by 3 and add to (2)

33126 zyx

232 zyx+

5107 yx (5)

Multiple (4) by -2 and add to (5)

0106 yx

5107 yx+

5x3y

3z

Example

Solve the system of linear equations

3411

742

3235

zyx

zyx

zyx (1)

(2)

(3)


3235 zyx14284 zyx+

1759 yx (4)


284168 zyx3411 zyx+

3159 yx (5)

Subtract (4) from (5)

3159 yx

1759 yx-

140

NO SOLUTION

Example

A basketball team scored 70 points in a game. The number of 2 point basketswas two more than the number of free throws, and the number of freethrows was one more than two times the number of three point baskets. What combination of scoring accounted for the teams’ 70 points?

Let x represent the number of 2 point baskets, y represent the number of three point baskets, and z represent the number of free throws.

7032 zyx

2zx

12 yz

(1)

(3)

(2)

Substitute (2) into (1)

703)2(2 zyz

6633 zy (4)

Substitute (3) into (4)

66)12(33 yy

639 y 7y

15z 17x

The team made 17 two point baskets, 15 free throws, and 7 three points baskets

Partial Fractions - Example

Write the partial fraction decomposition of the following rational expression.

xx 14

72

Step 1: If the power in the numerator is greater than or equal to that in the denominator - divide

Step 2: Completely factor the denominator

)14(

7

14

72

xxxx

Step 3: decompose the rational expression

14)14(

7

x

B

x

A

xx

Partial Fractions - Example


xx 14

72

Step 4: multiple by the LCD

)14(14)14(

7

xxx

B

x

A

xxBxxA )14(7

Step 5: solve – here plug in x = 0 to solve for A, and x = 14 to solve for B

2

1 and

2

1 BA

xxxx 2

1

)14(2

1

14

72

Example


12

2

x

x

Since the power in the numerator is equal to that in the denominator, the first step is to divide (use long division)

1

11

1 22

2

xx

x

Factor the denominator

)1)(1(

11

12

2

xxx

x

Decompose

)1()1()1)(1(

1

x

B

x

A

xx

Example


12

2

x

x

Step 4: multiple by the LCD

)1)(1()1()1()1)(1(

1

xxx

B

x

A

xx)1()1(1 xBxA

Step 5: solve – here plug in x = -1 to solve for A, and x = 1 to solve for B

2

1 and

2

1 BA

)1(2

1

)1(2

11

12

2

xxx

x

Example – Repeated Linear Factors


12

322

xx

x

Factor

22 )1(

32

12

32

x

x

xx

x

Decompose – here we have arepeated linear factor

22 )1()1()1(

32

x

B

x

A

x

x

Multiple by the LCD

222

)1()1()1()1(

32

xx

B

x

A

x

x

BxAx )1(32

Solve – plug in x = 1 to get B, thenany x value to get A (2 is easy)

1 and 2 BA

22 )1(

1

)1(

2

12

32

xxxx

x

Example – Quadratic Factor


xxx

xx

23

2 1

Factor

)1(

12

2

xxx

xx

Decompose – here we have alinear factor & a quadratic factor

)1()1(

122

2

xx

CBx

x

A

xxx

xx

Multiple by the LCD

)1()1()1(

1 222

2

xxxxx

CBx

x

A

xxx

xx

CxBxxxAxx 222 )1(1

AxCAxBAxx )()(1 22

To solve for A plug in x = 0

1A

Coefficients for the x2 and x termsmust be equal

1 BA 2B

1CA 0C

xxx

x

xxx

xx 1

1

21223

2

Example


43

122

23

xx

xxx

Divide – degree of numeratoris greater than denominator

43

361

43

1222

23

xx

xx

xx

xxx

Factor the fraction & Decompose

14)1)(4(

36

43

362

x

B

x

A

xx

x

xx

x

Multiple by the LCD

)1)(4(14)1)(4(

36

xx

x

B

x

A

xx

x

)4()1(36 xBxAx

Solve – plug in x = -4 to get A, thenx = 1 to get B

5

3 and

5

27 BA

)1(5

3

)4(5

271

43

122

23

xxx

xx

xxx

Example


632

623

xxx

x

Factor

)2(3)2(

6

632

6223

xxx

x

xxx

x

)2)(3(

6

632

6223

xx

x

xxx

x

23)2)(3(

622

x

C

x

BAx

xx

x

)2)(3(23)2)(3(

6 222

xx

x

C

x

BAx

xx

x

)3()2)((6 2 xCxBAxx

Plug in x = -2 to solve for C

4C

Coefficients for the x2 and x termsmust be equal

124226 22 xBAxBxAxx

122)2()4(6 2 BxBAxAx

04 A 4A

12 BA 9B

2

4

3

94

632

6223

xx

x

xxx

x

-10 -5 0 5 10

-20

-15

-10

-5

0

5

10

Inequalities

Sketch the graph of the inequality1535 yx

First sketch the graph of 1535 yx

Since the inequality is greater thanuse a dashed line to sketch the

graph. If it was just greater than or equal to use a solid line

Test a point above and below the lineto determine where to shade the

solution.

-3 -1 1 3 50

4

8

Example

Sketch the graph of the inequality

9)4()1( 22 yx

First sketch the graph of

9)4()1( 22 yx

Since it is greater than or equal to use a solid line

Test a point inside and outsideof the circle to determine

where to shade the solution.

Example

Sketch the graph and label the vertices of the solution set of the following system of inequalities.

935

62

yx

yx

First graph the line x - 2y = -6

-10 -5 0 5 10

-2

0

2

4

6

8 62 yx


solution.

Example


935

62

yx

yx

First graph the line 5x - 3y = -9


solution.

-10 -5 0 5 10

-40

-20

0

20

40

935 yx

Example


935

62

yx

yx

-10 -5 0 5 10

-2

0

2

4

6

8 62 yx

-10 -5 0 5 10

-40

-20

0

20

40

935 yx

Example

-10 -5 0

-15

-10

-5

0

5

10 935 yx

62 yx

Example


0

1

32

42

y

x

xy

xy

First find the intersection points between the parabola and the line

3242 xx

0722 xx

Use quad formula

83.1or 83.3 xx

Intersection points are:(3.83, 10.66) and (-1.83, -0.66)

-5 0 5

-10

0

10

20

1x

32 xy

42 xy

(3.83, 10.66)

Find the vertices ofthe solution set

Find the intersection of x = -1 and the line

(-1, 1)(-1, 1)

Obviously (-1, 0) isa one of the vertices

Find the positivex intercept of the

parabola

(2, 0)

(2, 0)

0

1

32

42

y

x

xy

xy

Example

A person plans to invest up to $20,000 in two different interest-bearing accounts. Each account is to contain at least $5000. Moreover, the amountin one account should be at least twice the amount in the other one. Findand graph a system of inequalities to describe the various amounts that

can be deposited in each account.

Let x represent the amount in one account and y the amount in the other.

20000 yx

xy 2

5000x

5000y

0 5000 10000 15000 200000

5000

10000

15000

20000

250005000x

5000y

xy 2

20000 yx

20000 yx

xy 2

5000x

5000y

Documents

Chapter 7 Review - Precalculus