Linear Equations Inequalities and Applications

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Math 1 - Chapter IV

Linear Equations, Inequalities, and Applications

• 4-1 Linear Equations in One Variable

• 4-2 Formulae

• 4-3 Applications

• 4-5 Linear Inequalities in One Variable

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2-1 Linear Equations in One Variable

A linear equation in one variable can be written in the form Ax + B = C, where A,B, and C are real numbers with A 0.

• Example 1: x + 4 = -2 Note: x + 4 by itself is called an algebraic expression.

• Example 2: 2k + 5 = 10 Note: A linear equation is also called a first degree equation since the highest power of x is 1.

• The following are called non-linear equations: y = 3x2 +10

2x

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Decide whether a number is a solution of a linear equation.

• Example 3: 8 is a solution of the equation x – 3 = 5

An equation is solved by finding its “solution set”.

The solution set of x – 3 = 5 is {8}.

Note: Equivalent equations are equations that have the same solution set:

• Example 4: 5x +2 = 17, 5x =15 and x =3 are equivalent.

The solution set is {3}.

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Addition Properties of Equality • For all real numbers A,B, and C, the

equations A = B and A + C = B + C are equivalent.

Note: The same number can be added to each side of an expression without changing its solution set.

• Example 5: Solution of the equation x – 3 = 5

x – 3 +3 = 5 + 3 x = 8

The solution set of x – 3 = 5 is {8}. • Example 6: 5x +2 = 17, 5x + 2 + (-2) =17 + (-2) and 5x =15

are equivalent. The solution set is {3}.

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Multiplication Properties of Equality

• For all real numbers A,B, and for

C 0, the equations A = B and AC = BC are equivalent.

Note: Each side of an equation may be multiplied by the same non-zero number without changing its solution set.

• Example 7: Solution of the equation

The solution set of 5x = 15 is {3}.

5x = 15

1 1( ) 5 ( ) 155 5

x = 3

x

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Example 8: Solve: -7 + 3x – 9x = 12x -5

-7 + 7 –6x = 12x –5 + 7

-6x –12x = 12x –12x +2

-18x = 2

The solution set is { }.

1 1( ) (-18) ( ) 2

18 181

x = -9

x

1-

9

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Solving a Linear Equation in One Variable:

• Step 1: Clear fractions. (Eliminate any fractions by multiplying each side by the least common denominator.)

• Step 2: Simplify each side separately. (Use the distributive property to clear parenthesis and combine like terms.)

• Step 3: Isolate the variable terms on the left side of the equal sign. (Use the addition property to get all the variable terms on the left side and all the numbers on the right side.)

• Step 4: Isolate the variable. (Use the multiplication property to get an equation where the coefficient of the variable is 1.)

• Step 5: Check. (Substitute the proposed solution into the original equation and verify both sides of the equal sign are equivalent.)

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Example 8: Solve: 6 – (4 +x) = 8x – 2(3x + 5)

Step 1: Does not applyStep 2: 6 – 4 –x = 8x – 6x –10

-x +2 = 2x – 10Step 3: -x + (-2x) +2 +(-2) = 2x + (-2x) –10 + (-2)

-3x = -12 Step 4:

Step 5: 6 – (4 + 4) = 8(4) – 2(3(4) + 5) 6 – 8 = 32 - 34 -2 = -2

The solution set is {4}.

1 1( ) (-3) ( ) (-12)

3 3 x = 4

x

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Example 9: Solve:

Step 1:

Step 2:

Step 3:

Step 4:

Step 5: The solution set is {-1}.

1 3 1

2 4 21 3 1

4( ) 4( )2 4 2

1 3 14( ) 4( ) 4( )

2 4 2 2( 1) ( 3) 2

2 2 3 2

3 5 2

3 5 ( 5) 2 ( 5)

3 3

x x

x x

x x

x x

x x

x

x

x

1 1

( ) 3 ( )( 3)3 3

1

1 1 1 3 1 2 1 1 1:

2 4 2 4 2 2 2

x

x

Check

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Identification of Conditional Equations, Contradictions, and Identities

• Some equations that appear to be linear have no solutions, while others have an infinite number of solutions.

Note: Identification requires knowledge of the information in the table.

Type of Equation

Number of Solutions

Indication when solving

Conditional One Final line is: x = a number

Contradiction None; solution set 0

or null

Final line is false, such

as 0=1

Identity Infinite; solution set

is all real numbers

Final line is true, such as

0=0

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Example 9: Solve each equation, Decide whether it is conditional, identity, or contradiction.

A) 5x – 9 = 4(x-3)5x – 9 = 4x –12 x = -3 conditional The solution set is {-3}

B) 5x –15 = 5(x-3)5x – 15 = 5x –15 0 = 0 identity The solution set is {all real numbers}

C) 5x –15 = 5(x-4)5x – 15 = 5x –20 0 = -5 contradiction The solution set is {} or 0 (null set)

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4-2 Formulae

A “mathematical model” is an equation or inequality that describes a real situation. Models for many applied problems already exist and are called “formulas”. A “formula” is a mathematical equation in which variables are used to describe a relationship.

Common formulae that will be used are:d = rt distance = (rate)(time) I = prt Interest = (principal)(rate)(time)P = 2L + 2 W Perimeter = 2(Length) + 2(Width)

Additional formulae are included on the inside covers of your text.

amount100% 100%

base

apercent percent

b

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4-2 Formulae

Solving for a specified variable: Step 1: Get all terms containing the specified variable on the left side of the equal sign and all other terms on the right side. Step 2: If necessary, use the distributive property to combine the terms with the specified variable. Step 3: Isolate the variable. (Use the multiplication (division) property to get an equation where the coefficient of the variable is 1.)

• Example 1: Solve m = 2k + 3b for k2k = m –3b k = (m – 3b)/2

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4-2 Formulae

• Example 2: Solve

• Example 3: Solve

1( 3) for

2y x x

1 3( )

2 21 3

( )2 2

1 32( ( )) 2( )

2 22 3

y x

x y

x y

x y

2 2 2 for A HW LW LH W 2 2 2

(2 2 ) 2

( 2 )

(2 2 )

HW LW A LH

W H L A LH

A LHW

H L

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4-2 Formulae

• Example 4: Given a distance of 500 miles and a rate (speed) of 25 mph, find the time for the trip.

• Example 5: A 20 oz mixture of gasoline and oil contains 1 oz of oil. What percent of the mixture is oil?

solve for d rt t

rt d

dt

r

500 20

25

d milest hours

milesrhour

100%amount

percentbase

1

100% .05 100% 5%20

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4-3 Applications of Linear Equations

Translating words to mathematical expressions.Note: There are usually key words or phrases that translate into mathematical expressions.

Translating from Words to Mathematical Expressions

x and y represent numbersVerbal Expression Mathematical Expression

AdditionsThe sum of a number and 7 x + 7

6 more than a number x + 63 plus a number 3 + x

24 added to a number x + 24A number increased by 5 x + 5The sum of two numbers x + y

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x and y represent the numbersVerbal Expression Mathematical Expression

Subtractions2 less than a number x - 2

3 minus a number 3 - xA number decreased by 5 x - 5

A number subtracted from 10 10 - xThe difference between two numbers x - y

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x and y represent the numbersVerbal Expression Mathematical Expression

Multiplication2 times a number 2 x or 2x

A number multiplied by 6 x 6 or 6x

Twice a number 2 x or 2xThe product of two numbers x y or xy

2 of a number

3

2 2 x or x

3 3

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4-3 Applications of Linear EquationsTranslating words in to equations.Note: The symbol for equality, = , is often indicated by the word “is”.

Verbal Sentence Mathematical Equation

Twice a number, decreased by 3, is 42 2 x - 3 = 42

The product of a number and 12, decreased by 7, is 105 12x - 7 = 105

The quotient of a number and the number plus 4 is 28

The quotient of a number and 4, plus the number, is 10

Translating from Words to Mathematical Equations

284

x

x

104

xx

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Distinguishing between expressions and equations.Note: An expression is a phrase. An equation includes the = symbol and translates the sentence.

• Example 1: Decide whether each is an expression or an equation.A) 5x – 3(x+2) = 7 Equation

B) 5x – 3(x+2) Expression

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Solving Applied Problems.Note: The following six steps are useful in developing a technique for solving applied problems.

• Step 1: Read the problem carefully until you understand what is given and what is to be found.

• Step 2: Assign a variable to represent the unknown value, using diagrams or tables as needed.

• Step 3: Write an equation using the variable expressions.• Step 4: Solve the equation. (Isolate the variable on the left side

of the equals sign) • Step 5: State the answer with appropriate units. (Does it seem

reasonable?)• Step 6: Check the answer.

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Solving Applied Problems.At the end of the 1999 baseball season, Sammy Sosa and Mark McGwire had a lifetime total of 858 homeruns. McGwire had 186 more than Sosa. How many runs did each player have?

• Step 1: Total runs = 858 Sosa has 186 less than McGwire

• Step 2: Let x = Sosa’s runs, then x + 186 = McGwire’s runs

• Step 3: x + (x + 186) = 858• Step 4: 2x + 186 = 858

2x = 672 x = 336 x + 186 = 522

• Step 5: Sosa had 336 homeruns. McGwire had 522 homeruns.• Step 6: 336 + 522 = 858

858 = 858Checked:

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Solving Applied Problems.A woman has $34,000 to invest. She invests some at 17% and the balance at 20%. Her total annual interest income is $6425. Find the amount invested at each rate.

• Step 1: Total investment = $34,000 Total interest earned is $6245

• Step 2: Let x = amount at 17%, then $34,000 - x = amount at 20%

• Step 3: .17x + .20(34,000 - x) = 6245• Step 4: .17x + 6800 - .20x = 6245

-.03x = -555 x = 18,500 34,000 – 18,500 = 15,500

• Step 5: $18,500 invested at 17% $15,500 invested at 20%• Step 6: .17(18,500) + .20(15,5000 = 6245

3145 + 3100 = 6245 6245 = 6245

Checked:

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Solving Applied Problems.How many pounds of candy worth $8 per lb should be mixed with 100 lb of candy worth $4 per lb to get a mixture that can be sold for $7 per lb?

• Step 1: Total mixture should sell for $7/lb. Given 100 lbs at $4/lb and an unknown amount at $8/lb.

• Step 2: Let x = amount at $8 then x + 100 should sell at $7/lb

• Step 3: 8x + 4(100) = 7(x + 100)• Step 4: 8x + 400 = 7x + 700

x = 300

• Step 5: 300 lb at $8/lb • Step 6: 8(300) + 4(100) = 7(300 + 100)

2400 + 400 = 7(400) 2800 = 2800

Checked:

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Solving Applied Problems.How much water must be added to 20 L of 50% antifreeze solution to reduce it to a 40% antifreeze solution?

• Step 1: Total mixture should be 40% antifreeze. Given 20 L at 50% produces 10 L of water and 10 L of antifreeze. Add an unknown amount of water.

• Step 2: Let x = amount of water added, then x + 10 will be the water in the final solution and the amount of antifreeze is 10 L.

• Step 3:

• Step 4:

• Step 5: 5 L of water • Step 6: Check:

• Checked:

amount100% 100%

base

apercent percent

b

10 L of antifreeze100% 40

10 L of antifreeze + (x+10) of water

1000 40(10 ( 10))

1000 400 40 400

40 200

5

x

x

x

x

10 L of antifreeze100% 40%

10 L of antifreeze + (5+10) of water

10100% 40%

2540% 40%

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4-5 Linear Inequalities in One Variable

Interval Notation is used to write solution sets of inequalities. Note: A parenthesis is used to indicate an endpoint in not included. A square bracket indicates the endpoint is included.

Type of Interval Set Interval Notation Graph

{x a x } (a, )

{x a x b} (a, b)

{x x b} (-, b)

{x x is a real number} (-, )

Interval Notation

Open Interval

(a

)b

(a

)b)b

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{x a x} [a, )

{x a x b} (a, b]

{x a x b} [a, b)

{x x b} (-, b]

Interval Notation

Half-open Interval

[a

]b

]b

(a

)b

[a

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Closed Interval {x a x b} [a, b]

Interval Notation

]b

[a

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A linear inequality in on variable can be written in the form Ax + B = C, where A, B, and C are real numbers with A 0. Note: The next examples include definitions and rules for < , >, , and .

• Examples of linear inequalities:

x + 5 < 2 x – 3 > 5 2k +4 10

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Solving linear inequalities using the Addition Property:

For all real numbers A, B, and C, the inequalities A < B and A + C < B + Ca are equivalent.Note: As with equations, the addition property can be used to add negative values or to subtract the same number from each side .

• Example 1: Solve k – 5 > 1

k – 5 +5 > 1 + 5k > 6

Solution set: (6,)• Example 2: Solve 5x + 3 4x – 1 and graph the solution

set.5x – 4x -1 –3x -4

Solution set: [-4,)

[-4

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Solving linear inequalities using the Multiplication Property:For all real numbers A, B, and C, with C 0 ,1) if C > 0, then the inequalities A < B and AC < BC are equivalent.2) if C < 0, then the inequalities A < B and AC > BC are equivalent.

Note: Multiplying or Dividing by a negative number requires the inequality sign be reversed.

• Example 1: Solve -2x < 10 x > -5

Solution set: (-5,)• Example 2: Solve 2x < -10

x < -5 Solution set: (-,-5)

Note: The first example requires a symbol change because both sides are multiplied by (-1/2). The second example does not because both sides are multiplied by (1/2)

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Solving linear inequalities using the Multiplication Property:

• Example 3: Solve –9m < -81 and graph the solution setm > 9

Solution set: (9, ) (9

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Solving a Linear Inequality: Step 1: Simplify each side separately. If necessary, use the distributive property to clear parenthesis and combine like terms.

Step 2: Use the addition property to get all terms containing the specified variable on the left side of the inequality sign and all other terms on the right side.

Step 3: Isolate the variable. (Use the multiplication (division) property to get an inequality where the coefficient of the variable is 1.)

• Example 4: Solve 6(x-1) + 3x -x –3(x + 2) and graph the solution setStep 1: 6x-6 + 3x -x –3x – 6

9x - 6 –4x – 6Step 2: 13x 0Step 3: x 0

Solution set: [0, )

[0

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Solving a Linear Inequality: • Example 5: Solve and graph the

solution set

Step 1:

Step 2:

Step 3:

Solution set: [ , )[

-13/2

1 3( 3) 2 ( 8)

4 4m m

1 3 32 6

4 4 41 11 3 24

4 4 4 41 3 24 11

4 4 4 42 13

4 413

2

m m

m m

m m

m

m

13

2

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Solving applied problems using linear inequalities:

• Example 6: Solve 5 < 3x – 4 < 9 and graph the solution set

5 + 4 < 3x < 9 + 4

9 < 3x < 13

3 < x < (13/3)

Solution set: ( 3, )

13

2 )(3 13

2

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Solving applied problems using linear inequalities: Note: Expressions for inequalities sometimes appear as indicated in the table:

• Example 7: Teresa has been saving dimes and nickels. She has three times as many nickels as dimes and she has at least 48 coins. What is the smallest number of nickels she might have?

Let: x equal the number of dimesThen: 3x is the number of nickelsand: 3x + x 48

4x 48 x 12 36 nickels

Word Expression Interpretation

a is at least b

a is no less than b

a is at most b

a is no more than b

a ba ba ba b

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Solving applied problems using linear inequalities:

• Example 8: Michael has scores of 92, 90, and 84 on his first three tests. What score must he get on the 4th test in order to maintain an average of at least 90?

Let: x equal the 4th score

92 90 84: 90

4266

904

266 360

360 266

94

xThen

x

x

x

x

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Linear Equations Inequalities and Applications