1. 3-6 Solving Equations Containing Integers Warm Up Problem of the Day
Course 2
Warm Up
Problem of the Day
Lesson Presentation

2. 3-6 Solving Equations Containing Integers Warm Up
Course 2
3-6
Solving Equations Containing Integers
Warm Up
Use mental math to find each solution.
y = 15
2. x ÷ 9 = 9
3. 6x = 24
4. x – 12 = 30
y = 8
x = 81
x = 4
x = 42

3. 3-6 Solving Equations Containing Integers Problem of the Day
Course 2
3-6
Solving Equations Containing Integers
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

4. 3-6 Learn to solve one-step equations with integers.
Course 2
3-6
Solving Equations Containing Integers
Learn to solve one-step equations with integers.

5. 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. + + + – – –
3 + (–3) = 0
a + (–a) = 0

6. 3-6 Solving Equations Containing Integers Helpful Hint 3 + (–3) = 0
Course 2
3-6
Solving Equations Containing Integers
3 + (–3) = 0
3 is the opposite of –3.
Helpful Hint

7. Additional Example 1A: Solving Addition and Subtraction Equations
Course 2
3-6
Solving Equations Containing Integers
Additional Example 1A: Solving Addition and Subtraction Equations
Solve. Check each answer.
A. –6 + x = –7
–6 + x = –7
Add 6 to both sides to
isolate the variable.
x = –1
Check
–6 + x = –7
Substitute –1 for x in the
original equation.
–6 + (–1) = –7 ?
–7 = –7 ? 
True. –1 is the solution to
–6 + x = –7.

8. Additional Example 1B: Solving Addition and Subtraction Equations
Course 2
3-6
Solving Equations Containing Integers
Additional Example 1B: Solving Addition and Subtraction Equations
Solve. Check the answer.
B. p + 5 = –3
p + 5 = –3
+ (–5) +(–5)
Add –5 to both sides to
isolate the variable.
p = –8
Check
p + 5 = –3
Substitute –8 for p in the
original equation.
–8 + 5 = –3 ?
– 3 = –3 ? 
True. –8 is the solution to
p + 5 = –3.

9. Additional Example 1C: Solving Addition and Subtraction Equations
Course 2
3-6
Solving Equations Containing Integers
Additional Example 1C: Solving Addition and Subtraction Equations
Solve. Check the answer.
C. y – 9 = –40
y – 9 = –40
Add 9 to both sides to
isolate the variable.
y = –31
Check
y – 9 = –40
Substitute –31 for y in the
original equation.
–31 – 9 = –40 ?
–40 = –40 ? 
True. –31 is the solution to
y – 9 = –40.

10. Insert Lesson Title Here
Course 2
3-6
Solving Equations Containing Integers
Insert Lesson Title Here
Try This: Example 1A
Solve. Check each answer.
A. –3 + x = –9
–3 + x = – 9
Add 3 to both sides to
isolate the variable.
x = –6
Check
–3 + x = –9
Substitute –6 for x in the
original equation.
–3 + (–6) = –9 ?
–9 = –9 ? 
True. –6 is the solution to
–3 + x = –9.

11. Insert Lesson Title Here
Course 2
3-6
Solving Equations Containing Integers
Insert Lesson Title Here
Try This: Example 1B
Solve. Check the answer.
B. q + 2 = –6
q + 2 = –6
+(–2) +(–2)
Add –2 to both sides to
isolate the variable.
q = –8
Check
q + 2 = –6
Substitute –8 for q in the
original equation.
–8 + 2 = –6 ?
–6 = –6 ? 
True. –8 is the solution to
q + 2 = –6.

12. Insert Lesson Title Here
Course 2
3-6
Solving Equations Containing Integers
Insert Lesson Title Here
Try This: Example 1C
Solve. Check the answer.
C. y – 7 = –34
y – 7 = –34
Add 7 to both sides to
isolate the variable.
y = –27
Check
y – 7 = –34
Substitute –27 for y in the
original equation.
–27 – 7 = –34 ?
–34 = –34 ? 
True. –27 is the solution to
y – 7 = –34.

13. Additional Example 2A: Multiplication and Division Equations
Course 2
3-6
Solving Equations Containing Integers
Additional Example 2A: Multiplication and Division Equations
Solve. Check each answer. b –5 A.
= 6 b –5
= 6 b –5
(–5)
= (–5)6
Multiply both sides by –5 to
isolate the variable.
b = –30

14. Additional Example 2A Continued
Course 2
3-6
Solving Equations Containing Integers
Additional Example 2A Continued
Check b –5
= 6
Substitute –30 for b in the
original equation.
–30 ?
= 6
– 5
True. –30 is the solution to
6 = 6  b –5
= 6.

15. Additional Example 2B: Solving Multiplication and Division Equations
Course 2
3-6
Solving Equations Containing Integers
Additional Example 2B: Solving Multiplication and Division Equations
Solve. Check each answer.
B. –400 = 8y
–400 = 8y
–400 = 8y
Divide both sides by 8 to
isolate the variable.
–50 = y

16. Additional Example 2B: Solving Multiplication and Division Equations
Course 2
3-6
Solving Equations Containing Integers
Additional Example 2B: Solving Multiplication and Division Equations
Check.
–400 = 8y
Substitute –50 for y in the
original equation. ?
–400 = 8(–50)
True. –50 is the solution to
–400 = 8y.
–400 = –400 

17. Insert Lesson Title Here
Course 2
3-6
Solving Equations Containing Integers
Insert Lesson Title Here
Try This: Example 2A
Solve. Check each answer. c 4 A.
= –24 c 4
= –24 c 4 4
= 4(–24)
Multiply both sides by 4 to
isolate the variable.
c = –96

18. Insert Lesson Title Here
Course 2
3-6
Solving Equations Containing Integers
Insert Lesson Title Here
Try This: Example 2A Continued
Check c 4
= –24
Substitute –96 for c in the
original equation.
–96 ?
= –24 4
True. –96 is the solution to
–24 = –24  c 4
= –24.

19. Insert Lesson Title Here
Course 2
3-6
Solving Equations Containing Integers
Insert Lesson Title Here
Try This: Example 2B
Solve. Check each answer.
B. –200 = 4x
–200 = 4x
–200 = 4x
Divide both sides by 4 to
isolate the variable.
–50 = x

20. Insert Lesson Title Here
Course 2
3-6
Solving Equations Containing Integers
Insert Lesson Title Here
Try This: Example 2B Continued
Check.
–200 = 4x
Substitute –50 for x in the
original equation. ?
–200 = 4(–50)
True. –50 is the solution to
–200 = 4x.
–200 = –200 

21. Additional Example 3: Business Application
Course 2
3-6
Solving Equations Containing Integers
Additional Example 3: Business Application
In 2003, a manufacturer made a profit of $300 million. This amount was $100 million more than the profit in What was the profit in 2002?
Let p represent the profit in 2002 (in millions of dollars).
Profit in 2003 = p + 100
Profit in 2003 = $300 million
p = 300
–100 –100
p = 200
The profit in 2002 was $200 million.

22. Insert Lesson Title Here
Course 2
3-6
Solving Equations Containing Integers
Insert Lesson Title Here
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?
Let x represent the money made last year.
This year = x + 125
This year = $243
x = 243
–125 –125
x = 118
The class earned $118 last year.

23. Insert Lesson Title Here
Course 2
3-6
Solving Equations Containing Integers
Insert Lesson Title Here
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?
100 4 y 7
–49
–93
$39 million