chapter 9 equations, inequalities and problem solving

Chapter 9

Equations, Inequalities and Problem Solving

Martin-Gay, Introductory Algebra, 3ed 2Martin-Gay, Developmental Mathematics 2

9.1 – The Addition Property of Equality

9.2 – The Multiplication Property of Equality

9.3 – Further Solving Linear Equations

9.4 – Introduction to Problem Solving

9.5 – Formulas and Problem Solving

9.6 – Percent and Mixture Problem Solving

9.7 – Solving Linear Inequalities

Chapter Sections

§ 9.1

The Addition Property of Equality


Linear Equations

Linear equation in one variable

can be written in the form ax + b = c, a 0

Equivalent equations

are equations with the same solutions in the form of

variable = number, or

number = variable


Addition Property of Equality

a = b and a + c = b + c are equivalent equations

Addition Property of Equality

z = – 16 (Simplify both sides)

Example

8 + (– 8) + z = – 8 + – 8 (Add –8 to each side)

8 + z = – 8a.)


Example

Solving Equations

4p – 11 – p = 2 + 2p – 20

3p – 11 = 2p – 18 (Simplify both sides)

p = – 7 (Simplify both sides)

p – 11 = – 18 (Simplify both sides)

3p + (– 2p) – 11 = 2p + (– 2p) – 18 (Add –2p to both sides)

p – 11 + 11 = – 18 + 11 (Add 11 to both sides)


Example

Solving Equations

5(3 + z) – (8z + 9) = – 4z

15 + 5z – 8z – 9 = – 4z (Use distributive property)

6 – 3z = – 4z (Simplify left side)

6 + z = 0 (Simplify both sides)

z = – 6 (Simplify both sides)

6 – 3z + 4z = – 4z + 4z (Add 4z to both sides)

6 + (– 6) + z = 0 +( – 6) (Add –6 to both sides)

§ 9.2

The Multiplication Property of Equality


Multiplication property of equalitya = b and ac = bc are equivalent equations

Multiplication Property of Equality

Example

– y = 8

y = – 8 (Simplify both sides)

(– 1)(– y) = 8(– 1) (Multiply both sides by –1)


Example

Solving Equations

9

5

7

1x

(Simplify both sides)9

35x

79

5

7

17

x (Multiply both sides by 7)


Example

Solving Equations

16x (Simplify both sides)

8 63

x

3

86

8

3

3

8x (Multiply both sides by fraction)


Example

Solving Equations

Recall that multiplying by a number is equivalent to dividing by its reciprocal

3z – 1 = 26

3z = 27 (Simplify both sides)

z = 9 (Simplify both sides)

3z – 1 + 1 = 26 + 1 (Add 1 to both sides)

(Divide both sides by 3)3

27

3

3

z


Example

Solving Equations

12x + 30 + 8x – 6 = 10

20x + 24 = 10 (Simplify left side)

20x = – 14 (Simplify both sides)

20

14

20

20

x(Divide both sides by 20)

10

7x (Simplify both sides)

20x + 24 + (– 24) = 10 + (– 24) (Add –24 to both sides)

§ 9.3

Further Solving Linear Equations


Solving Linear Equations

Solving linear equations in one variable1) Multiply to clear fractions

2) Use distributive property

3) Simplify each side of equation

4) Get all variable terms on one side and number terms on the other side of equation (addition property of equality)

5) Get variable alone (multiplication property of equality)

6) Check solution by substituting into original problem


301093 yy (Simplify)

7

7

7

21 y

(Simplify; divide both sides by 7)

y 3 (Simplify both sides)


Example3( 3) 2 6

5y y

(Multiply both sides by 5) 6255

)3(35

y

y

(Add –3y to both sides)30)3(109)3(3 yyyy

(Simplify; add –30 to both sides))30(307)30(9 y



Example

5x – 5 = 2(x + 1) + 3x – 7

5x – 5 = 2x + 2 + 3x – 7 (Use distributive property)

5x – 5 = 5x – 5 (Simplify the right side)

Both sides of the equation are identical. Since this equation will be true for every x that is substituted into the equation, the solution is “all real numbers.”



Example

Since no value for the variable x can be substituted into this equation that will make this a true statement, there is “no solution.”

3x – 7 = 3(x + 1)

3x – 7 = 3x + 3 (Use distributive property)

– 7 = 3 (Simplify both sides)

3x + (– 3x) – 7 = 3x + (– 3x) + 3 (Add –3x to both sides)

§ 9.4

An Introduction to Problem Solving


Strategy for Problem Solving

General Strategy for Problem Solving1) Understand the problem

• Read and reread the problem• Choose a variable to represent the unknown• Construct a drawing, whenever possible• Propose a solution and check

2) Translate the problem into an equation3) Solve the equation4) Interpret the result

• Check proposed solution in problem• State your conclusion


The product of twice a number and three is the same as the difference of five times the number and ¾. Find the number.

1.) Understand

Read and reread the problem. If we let

x = the unknown number, then “twice a number” translates to 2x,

“the product of twice a number and three” translates to 2x · 3,

“five times the number” translates to 5x, and

“the difference of five times the number and ¾” translates to 5x – ¾.

Finding an Unknown Number

Example

Continued


The product of

·

twice a number

2x

and 3

3

is the same as

=

5 times the number

5x

and ¾

¾

the difference of

–


Example continued

2.) Translate

Continued



Example continued3.) Solve

2x · 3 = 5x – ¾

6x = 5x – ¾ (Simplify left side)

x = – ¾ (Simplify both sides)

6x + (– 5x) = 5x + (– 5x) – ¾ (Add –5x to both sides)

4.) InterpretCheck: Replace “number” in the original statement of the problem with – ¾. The product of twice – ¾ and 3 is 2(– ¾)(3) = – 4.5. The difference of five times – ¾ and ¾ is 5(– ¾) – ¾ = – 4.5. We get the same results for both portions.

State: The number is – ¾.


A car rental agency advertised renting a Buick Century for $24.95 per day and $0.29 per mile. If you rent this car for 2 days, how many whole miles can you drive on a $100 budget?

1.) UnderstandRead and reread the problem. Let’s propose that we drive a total of 100 miles over the 2 days. Then we need to take twice the daily rate and add the fee for mileage to get 2(24.95) + 0.29(100) = 49.90 + 29 = 78.90. This gives us an idea of how the cost is calculated, and also know that the number of miles will be greater than 100. If we let

x = the number of whole miles driven, then

0.29x = the cost for mileage driven

Solving a Problem

Example

Continued


Solving a Problem

Example continued

2.) Translate

Continued

Daily costs

2(24.95)

mileage costs

0.29x

plus

+

is equal to

= 100

maximum budget


Solving a Problem

Example continued3.) Solve

Continued

2(24.95) + 0.29x = 100

49.90 + 0.29x = 100 (Simplify left side)

0.29x = 50.10 (Simplify both sides)

29.0

10.50

29.0

29.0

x(Divide both sides by 0.29)

x 172.75 (Simplify both sides)

(Subtract 49.90 from both sides)49.90 – 49.90 + 0.29x = 100 – 49.90


Solving a Problem

Example continued

4.) Interpret

Check: Recall that the original statement of the problem asked for a “whole number” of miles. If we replace “number of miles” in the problem with 173, then 49.90 + 0.29(173) = 100.07, which is over our budget. However, 49.90 + 0.29(172) = 99.78, which is within the budget.

State: The maximum number of whole number miles is 172.

§ 9.5

Formulas and Problem Solving


Formulas

A formula is an equation that states a known relationship among multiple quantities (has more than one variable in it)

A = lw (Area of a rectangle = length · width)

I = PRT (Simple Interest = Principal · Rate · Time)

P = a + b + c (Perimeter of a triangle = side a + side b + side c)

d = rt (distance = rate · time)

V = lwh (Volume of a rectangular solid = length · width · height)


Using Formulas

A flower bed is in the shape of a triangle with one side twice the length of the shortest side, and the third side is 30 feet more than the length of the shortest side. Find the dimensions if the perimeter is 102 feet.

1.) Understand

Read and reread the problem. Recall that the formula for the perimeter of a triangle is P = a + b + c. If we let

x = the length of the shortest side, then

2x = the length of the second side, and

x + 30 = the length of the third side

Example

Continued


Using Formulas

Example continued

2.) Translate

Continued

Formula: P = a + b + c Substitute: 102 = x + 2x + x + 30

3.) Solve102 = x + 2x + x + 30

102 = 4x + 30 (Simplify right side)

102 – 30 = 4x + 30 – 30 (Subtract 30 from both sides)

72 = 4x (Simplify both sides)

4

4

4

72 x (Divide both sides by 4)

18 = x (Simplify both sides)


Using Formulas

Example continued

4.) Interpret

Check: If the shortest side of the triangle is 18 feet, then the second side is 2(18) = 36 feet, and the third side is 18 + 30 = 48 feet. This gives a perimeter of P = 18 + 36 + 48 = 102 feet, the correct perimeter.

State: The three sides of the triangle have a length of 18 feet, 36 feet, and 48 feet.


It is often necessary to rewrite a formula so that it is solved for one of the variables.

This is accomplished by isolating the designated variable on one side of the equal sign.

Solving Formulas

Solving Equations for a Specific Variable1) Multiply to clear fractions

2) Use distributive to remove grouping symbols

3) Combine like terms to simply each side

4) Get all terms containing specified variable on the same time, other terms on opposite side

5) Isolate the specified variable


Solving Equations for a Specific Variable

Example

mnrT

mr

mnr

mr

T (Divide both sides by mr)

nmr

T (Simplify right side)

Solve for n.



Example

Solve for T.

PRTPPPA (Subtract P from both sides)

PRTPA (Simplify right side)

PR

PRT

PR

PA

(Divide both sides by PR)

TPR

PA

(Simplify right side)

A P PRT



Example

Solve for P.A P PRT

)1( RTPA (Factor out P from both terms on the right side)

RT

RTP

RT

A

1

)1(

1(Divide both sides by 1 + RT)

PRT

A

1(Simplify the right side)

§ 9.6

Percent and Mixture Problem Solving


Solving a Percent Problem

A percent problem has three different parts:

1. When we do not know the amount:n = 10% · 500

Any one of the three quantities may be unknown.

amount = percent · base

2. When we do not know the base:50 = 10% · n

3. When we do not know the percent:50 = n · 500


Solving a Percent Problem: Amount Unknown


What is 9% of 65?

n = 9% · 65

n = (0.09) (65)

n = 5.85

5.85 is 9% of 65


Solving a Percent Problem: Base Unknown


36 is 6% of what?

n = 6% ·36

36 = 0.06n36 0.06 =

0.06 0.06n

600 = n

36 is 6% of 600


Solving a Percent Problem: Percent Unknown


n = 14424

24 is what percent of 144?

24 = 144n

24 144 = 144 144

n

0.16 = n216 % = 3

n 224 is 16 % of 1443


Solving Markup Problems

Mark is taking Peggy out to dinner. He has $66 to spend. If he wants to tip the server 20%, how much can he afford to spend on the meal?

Let n = the cost of the meal.

Cost of meal n + tip of 20% of the cost = $66

100% of n + 20% of n = $66120% of n = $66

1.2 66n 1.2 66 1.2 1.2

n

55n Mark and Peggy can spend up to $55 on the meal itself.

Example


Solving Discount Problems

Julie bought a leather sofa that was on sale for 35% off the original price of $1200. What was the discount? How much did Julie pay for the sofa?

Julie paid $780 for the sofa.

Discount = discount rate list price

= 35% 1200= 420 The discount was $420.

Amount paid = list price – discount

= 1200 – 420

= 780

Example


Solving Increase Problems

The cost of a certain car increased from $16,000 last year to $17,280 this year. What was the percent of increase?

Amount of increase = original amount – new amount

The car’s cost increased by 8%.

amount of increase = original amount

Percent of increase

= 17,280 – 16,000 = 1280

amount of increasePercent of increase = original amount

1280= 16000

= 0.08

Example


Solving Decrease Problems

Patrick weighed 285 pounds two years ago. After dieting, he reduced his weight to 171 pounds. What was the percent of decrease in his weight?

Amount of decrease = original amount – new amount

Patrick’s weight decreased by 40%.

amount of decrease = original amount

Percent of decrease

= 285 – 171 = 114

amount of decreasePercent of decrease = original amount

114= 285

= 0.4

Example


Solving Mixture Problems

The owner of a candy store is mixing candy worth $6 per pound with candy worth $8 per pound. She wants to obtain 144 pounds of candy worth $7.50 per pound. How much of each type of candy should she use in the mixture?

1.) Understand

Let n = the number of pounds of candy costing $6 per pound.

Since the total needs to be 144 pounds, we can use 144 n for the candy costing $8 per pound.

Example

Continued



Example continued

2.) Translate

Continued

Use a table to summarize the information.

Number of Pounds Price per Pound Value of Candy$6 candy n 6 6n$8 candy 144 n 8 8(144 n)$7.50 candy 144 7.50 144(7.50)

6n + 8(144 n) = 144(7.5)

# of pounds of $6 candy

# of pounds of $8 candy

# of pounds of

$7.50 candy



Example continued

Continued

3.) Solve

6n + 8(144 n) = 144(7.5)

6n + 1152 8n = 1080

1152 2n = 1080

2n = 72

(Eliminate the parentheses)

(Combine like terms)

(Subtract 1152 from both sides)

n = 36 (Divide both sides by 2)

She should use 36 pounds of the $6 per pound candy.

She should use 108 pounds of the $8 per pound candy.

(144 n) = 144 36 = 108



Example continued

4.) Interpret

Check: Will using 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy yield 144 pounds of candy costing $7.50 per pound?

State: She should use 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy.

6(36) + 8(108) = 144(7.5)?

216 + 864 = 1080 ?

1080 = 1080?

§ 9.7

Solving Linear Inequalities


Linear Inequalities

A linear inequality in one variable is an equation that can be written in the form ax + b < c

• a, b, and c are real numbers, a 0• < symbol could be replaced by > or or


Graphing solutions to linear inequalities in one variable (using circles)

• Use a number line• Use a closed circle at the endpoint of a interval

if you want to include the point• Use an open circle at the endpoint if you DO

NOT want to include the point

7Represents the set {xx 7}

-4Represents the set {xx > – 4}

Graphing Solutions


Graphing solutions to linear inequalities in one variable (using interval notation)

• Use a number line

• Use a bracket at the endpoint of a interval if you want to include the point

• Use a parenthesis at the endpoint if you DO NOT want to include the point

7]

-4( Represents the set (– 4, )

Represents the set (– , 7]

Interval Notation

Linear Inequalities


Addition Property of Inequality• a < b and a + c < b + c are equivalent inequalities

Multiplication Property of Inequality• a < b and ac < bc are equivalent inequalities, if c is positive

• a < b and ac > bc are equivalent inequalities, if c is negative

Properties of Inequality


Solving linear inequalities in one variable1) Multiply to clear fractions

2) Use distributive property

3) Simplify each side of equation

4) Get all variable terms on one side and numbers on the other side of equation (addition property of equality)

5) Isolate variable (multiplication property of equality)



3x + 9 5(x – 1)

3x + 9 5x – 5 (Use distributive property on right side)

3x – 3x + 9 5x – 3x – 5 (Subtract 3x from both sides)

9 2x – 5 (Simplify both sides)

14 2x (Simplify both sides)

7 x (Divide both sides by 2)

9 + 5 2x – 5 + 5 (Add 5 to both sides)

Graph of solution (– ,7]7]


Example


7(x – 2) + x > – 4(5 – x) – 12 7x – 14 + x > – 20 + 4x – 12 (Use distributive property) 8x – 14 > 4x – 32 (Simplify both sides)8x – 4x – 14 > 4x – 4x – 32 (Subtract 4x from both sides) 4x – 14 > –32 (Simplify both sides)4x – 14 + 14 > –32 + 14 (Add 14 to both sides) 4x > –18 (Simplify both sides)

2

9x (Divide both sides by 4 and simplify)

-92

(Graph of solution ( ,)-92


Example


A compound inequality is two inequalities joined together.

To solve the compound inequality, perform operations simultaneously to all three parts of the inequality (left, middle and right).

Compound Inequalities

0 4(5 – x) < 8


0 4(5 – x) < 8

0 20 – 4x < 8 (Use the distributive property)

3 5](

Graph of solution (3,5]

– 20 – 4x < – 12 (Simplify each part)

5 x > 3 (Divide each part by –4)

Remember that the sign direction changes when you divide by a number < 0!

0 – 20 20 – 20 – 4x < 8 – 20 (Subtract 20 from each part)

Solving Compound Inequalities

Example

chapter 9 equations, inequalities and problem solving

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