3-6 Solving Equations Containing Integers

Course 2

Warm Up

Problem of the Day

Lesson Presentation

Warm UpUse mental math to find each solution.

1. 7 + y = 152. x ÷ 9 = 93. 6x = 244. x – 12 = 30

y = 8x = 81x = 4

Course 2

3-6 Solving Equations Containing Integers

x = 42

Problem of the Day

Zelda sold her wet suit to a friend for $156. She sold her tank, mask, and snorkel for $85 less than she sold her wet suit. She bought a used wet suit for $80 and a used tank, mask, and snorkel for $36. If she started with $0, how much money does she have left?$111

Course 2

3-6 Solving Equations Containing Integers

Learn to solve one-step equations with integers.

Course 2

3-6 Solving Equations Containing Integers

Course 2

3-6 Solving Equations Containing Integers

When you are solving equations with integers, the goal is the same as with whole numbers—isolate the variable on one side of the equation. One way to isolate the variable is to add opposites. Recall that the sum of a number and its opposite is 0.

+ + +0

– – –

3 + (–3) = 0

a + (–a) = 0

Course 2

3-6 Solving Equations Containing Integers

3 + (–3) = 0

3 is the opposite of –3.

Helpful Hint

Solve. Check each answer.

Additional Example 1A: Solving Addition and Subtraction Equations

Course 2

3-6 Solving Equations Containing Integers

A. –6 + x = –7

–6 + x = –7+ 6 + 6

x = –1Check

–6 + x = –7

–6 + (–1) = –7?

Add 6 to both sides to isolate the variable.

Substitute –1 for x in theoriginal equation.

True. –1 is the solution to–6 + x = –7.

–7 = –7?

Additional Example 1B: Solving Addition and Subtraction Equations

Course 2

3-6 Solving Equations Containing Integers

Solve. Check the answer.

B. p + 5 = –3

p + 5 = –3

+ (–5) +(–5)

p = –8Check

–8 + 5 = –3?

Add –5 to both sides to isolate the variable.

Substitute –8 for p in theoriginal equation.

True. –8 is the solution top + 5 = –3.

p + 5 = –3

– 3 = –3 ?

Additional Example 1C: Solving Addition and Subtraction Equations

Course 2

3-6 Solving Equations Containing Integers

Solve. Check the answer.

C. y – 9 = –40

y – 9 = –40+ 9 + 9

y = –31Check

–31 – 9 = –40?

Add 9 to both sides to isolate the variable.

Substitute –31 for y in theoriginal equation.

True. –31 is the solution toy – 9 = –40.

y – 9 = –40

–40 = –40?

Try This: Example 1A

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Course 2

3-6 Solving Equations Containing Integers

Solve. Check each answer.

A. –3 + x = –9

–3 + x = – 9+ 3 + 3

x = –6Check

–3 + x = –9

–3 + (–6) = –9?

Add 3 to both sides to isolate the variable.

Substitute –6 for x in theoriginal equation.

True. –6 is the solution to–3 + x = –9.

–9 = –9 ?

Try This: Example 1B

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Course 2

3-6 Solving Equations Containing Integers

Solve. Check the answer.

B. q + 2 = –6

q + 2 = –6

+(–2) +(–2)

q = –8Check

–8 + 2 = –6?

Add –2 to both sides to isolate the variable.

Substitute –8 for q in theoriginal equation.

True. –8 is the solution toq + 2 = –6.

q + 2 = –6

–6 = –6?

Try This: Example 1C

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Course 2

3-6 Solving Equations Containing Integers

Solve. Check the answer.

C. y – 7 = –34

y – 7 = –34

+7 +7

y = –27Check

–27 – 7 = –34?

Add 7 to both sides to isolate the variable.

Substitute –27 for y in theoriginal equation.

True. –27 is the solution toy – 7 = –34.

y – 7 = –34

–34 = –34?

Solve. Check each answer.

Additional Example 2A: Multiplication and Division Equations

Course 2

3-6 Solving Equations Containing Integers

A. b–5

= 6

b–5

= 6

b–5 = (–5)6(–5)

b = –30

Multiply both sides by –5 toisolate the variable.

Additional Example 2A Continued

Course 2

3-6 Solving Equations Containing Integers

Check

b–5

= 6

–30– 5

= 6?

6 = 6

Substitute –30 for b in theoriginal equation.

True. –30 is the solution tob–5

= 6.

Additional Example 2B: Solving Multiplication and Division Equations

Course 2

3-6 Solving Equations Containing Integers

Solve. Check each answer.

B. –400 = 8y

–400 = 8y

–400 = 8y

8 8

–50 = y

Divide both sides by 8 to isolate the variable.

Check.

Additional Example 2B: Solving Multiplication and Division Equations

Course 2

3-6 Solving Equations Containing Integers

–400 = 8y

–400 = 8(–50)?

–400 = –400

Substitute –50 for y in theoriginal equation.

True. –50 is the solution to–400 = 8y.

Try This: Example 2A

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Course 2

3-6 Solving Equations Containing Integers

Solve. Check each answer.

A. c4

= –24

c4

= –24

c4 = 4(–24)4

c = –96

Multiply both sides by 4 toisolate the variable.

Try This: Example 2A Continued

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Course 2

3-6 Solving Equations Containing Integers

Check

c4

= –24

–964

= –24?

–24 = –24

Substitute –96 for c in theoriginal equation.

True. –96 is the solution toc4

= –24.

Try This: Example 2B

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Course 2

3-6 Solving Equations Containing Integers

Solve. Check each answer.

B. –200 = 4x

–200 = 4x

–200 = 4x4 4

–50 = x

Divide both sides by 4 to isolate the variable.

Try This: Example 2B Continued

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Course 2

3-6 Solving Equations Containing Integers

Check.

–200 = 4x

–200 = 4(–50)?

–200 = –200

Substitute –50 for x in theoriginal equation.

True. –50 is the solution to–200 = 4x.

In 2003, a manufacturer made a profit of $300 million. This amount was $100 million more than the profit in 2002. What was the profit in 2002?

Additional Example 3: Business Application

Course 2

3-6 Solving Equations Containing Integers

Let p represent the profit in 2002 (in millions of dollars).

Profit in 2003 = p + 100

Profit in 2003 = $300 million

p + 100 = 300–100 –100

p = 200

The profit in 2002 was $200 million.

Try This: Example 3

This year the class bake sale made a profit of $243. This was an increase of $125 over last year. How much did they make last year?

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Course 2

3-6 Solving Equations Containing Integers

This year = x + 125

This year = $243

x + 125 = 243–125 –125

x = 118

The class earned $118 last year.

Let x represent the money made last year.

Lesson Quiz

Solve. Check your answer.

1. –8y = –800

2. x – 22 = –18

3. – = 7

4. w + 72 = –21

5. Last year a phone company had a loss of $25 million. This year the loss is $14 million more last year. What is this years loss?

4

100

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–49

–93

Course 2

3-6 Solving Equations Containing Integers

y7

$39 million