1. Solving Addition and Subtraction Equations 1-3
Warm Up
Problem of the Day
Lesson Presentation
Pre-Algebra

2. 1-3 Solving Addition and Subtraction Equations Warm Up
Pre-Algebra
Warm Up
Write an algebraic expression for each word phrase.
1. a number x decreased by 9
2. 5 times the sum of p and 6
3. 2 plus the product of 8 and n
4. the quotient of 4 and a number c
x  9
5(p + 6)
2 + 8n 4 c

3. Problem of the Day
Janie’s horse refused to do 5 jumps today and cleared 14 jumps. Yesterday, the horse cleared 9 more jumps than today. He won 3 first place ribbons. How many jumps did the horse clear in the two-day jumping event? 37

4. Learn to solve equations using addition and subtraction.

5. Vocabulary equation solve solution inverse operation
isolate the variable
Addition Property of Equality
Subtraction Property of Equality

6. An equation uses an equal sign to show that two expressions are equal
An equation uses an equal sign to show that two expressions are equal. All of these are equations.
100 2
= 50
3 + 8 = 11
r + 6 = 14
24 = x – 7
To solve an equation, find the value of the variable that makes the equation true. This value of the variable is called the solution of the equation.

7. Determine which value of x is a solution of the equation.
Additional Example 1: Determining Whether a Number is a Solution of an Equation
Determine which value of x is a solution of the equation.
x + 8 = 15; x = 5, 7, or 23
Substitute each value for x in the equation.
x + 8 = 15 ?
5 + 8 = 15 ?
Substitute 5 for x.
13= 15 ? 
So 5 is not solution.

8. Additional Example 1 Continued
Determine which value of x is a solution of the equation.
x + 8 = 15; x = 5, 7, or 23
Substitute each value for x in the equation.
x + 8 = 15 ?
7 + 8 = 15 ?
Substitute 7 for x.
15= 15 ? 
So 7 is a solution.

9. Additional Example 1 Continued
Determine which value of x is a solution of the equation.
x + 8 = 15; x = 5, 7, or 23
Substitute each value for x in the equation.
x + 8 = 15 ?
= 15 ?
Substitute 23 for x.
31= 15 ? 
So 23 is not a solution.

10. Determine which value of x is a solution of the equation.
Try This: Example 1
Determine which value of x is a solution of the equation.
x – 4 = 13; x = 9, 17, or 27
Substitute each value for x in the equation.
x – 4 = 13 ?
9 – 4 = 13 ?
Substitute 9 for x.
5 = 13 ? 
So 9 is not a solution.

11. Try This: Example 1 Continued
Determine which value of x is a solution of the equation.
x – 4 = 13; x = 9, 17, or 27
Substitute each value for x in the equation.
x – 4 = 13 ?
17 – 4 = 13 ?
Substitute 17 for x.
13 = 13 ? 
So 17 is a solution.

12. Try This: Example 1 Continued
Determine which value of x is a solution of the equation.
x – 4 = 13; x = 9, 17, or 27
Substitute each value for x in the equation.
x – 4 = 13 ?
27 – 4 = 13 ?
Substitute 27 for x.
23 = 13 ? 
So 27 is not a solution.

13. Addition and subtraction are inverse operations, which means they “undo” each other.
To solve an equation, use inverse operations to isolate the variable. This means getting the variable alone on one side of the equal sign.

14. ADDITION PROPERTY OF EQUALITY
To solve a subtraction equation, like y  15 = 7, you would use the Addition Property of Equality.
ADDITION PROPERTY OF EQUALITY
Words
Numbers
Algebra
You can add the same number to both sides of an equation, and the statement will still be true.
2 + 3 = 5
x = y
+ 4
x = y
+ z
2 + 7 = 9

15. SUBTRACTION PROPERTY OF EQUALITY
There is a similar property for solving addition equations, like x + 9 = 11. It is called the Subtraction Property of Equality.
SUBTRACTION PROPERTY OF EQUALITY
Words
Numbers
Algebra
You can subtract the same number from both sides of an equation, and the statement will still be true.
4 + 7 = 11
x = y
 3
x = y
 z
4 + 4 = 8

16. Subtract 10 from both sides. 0 + n = 8 n = 8
Additional Example 2A: Solving Equations Using Addition and Subtraction Properties
Solve.
A n = 18
10 + n = 18
–10
–10
Subtract 10 from both sides.
0 + n = 8
n = 8
Identity Property of Zero: 0 + n = n.
Check
10 + n = 18 ?
= 18
18 = 18 ? 

17. Identity Property of Zero: p + 0 = p. Check p – 8 = 9 17 – 8 = 9
Additional Example 2B: Solving Equations Using Addition and Subtraction Properties
Solve.
B. p – 8 = 9
p – 8 = 9
+ 8
+ 8
Add 8 to both sides.
p + 0 = 17
p = 17
Identity Property of Zero: p + 0 = p.
Check
p – 8 = 9 ?
17 – 8 = 9
9 = 9 ? 

18. Identity Property of Zero: y + 0 = 0. Check 22 = y – 11 22 = 33 – 11
Additional Example 2C: Solving Equations Using Addition and Subtraction Properties
Solve.
C. 22 = y – 11
22 = y – 11
+ 11
+ 11
Add 11 to both sides.
33 = y + 0
33 = y
Identity Property of Zero: y + 0 = 0.
Check
22 = y – 11 ?
22 = 33 – 11
22 = 22 ? 

19. Subtract 15 from both sides. 0 + n = 14 n = 14
Try This: Example 2A
Solve.
A n = 29
15 + n = 29
–15
–15
Subtract 15 from both sides.
0 + n = 14
n = 14
Identity Property of Zero: 0 + n = n.
Check
15 + n = 29 ?
= 29
29 = 29 ? 

20. Identity Property of Zero: p + 0 = p. Check p – 6 = 7 13 – 6 = 7
Try This: Example 2B
Solve.
B. p – 6 = 7
p – 6 = 7
+ 6
+ 6
Add 6 to both sides.
p + 0 = 13
p = 13
Identity Property of Zero: p + 0 = p.
Check
p – 6 = 7 ?
13 – 6 = 7
7 = 7 ? 

21. Identity Property of Zero: y + 0 = 0. Check 44 = y – 23 44 = 67 – 23
Try This: Example 2C
Solve.
C. 44 = y – 23
44 = y – 23
+ 23
+ 23
Add 23 to both sides.
67 = y + 0
67 = y
Identity Property of Zero: y + 0 = 0.
Check
44 = y – 23 ?
44 = 67 – 23
44 = 44 ? 

22. = = + + Additional Example 3A Solve: x 34 16,550 x + 34 = 16,550 –34
A. Jan took a 34-mile trip in her car, and the odometer showed 16,550 miles at the end of the trip. What was the original odometer reading?
odometer reading at the beginning of the trip
miles traveled + =
odometer reading at the end of the trip + =
Solve: x 34
16,550
x + 34 = 16,550
–34
– 34
Subtract 34 from both sides.
x + 0 = 16,516
x = 16,516
The original odometer reading was 16,516 miles.

23. = = + + Additional Example 3B Solve: 895 n 1125 895 + n = 1125 –895
B. From 1980 to 2000, the population of a town increased from 895 residents to 1125 residents. What was the increase in population during that 20-year period?
initial population
increase in population + =
population after increase + =
Solve:
895 n
1125
895 + n = 1125
–895
– 895
Subtract 895 from both sides.
0 + n = 230
n = 230
The increase in population was 230.

24. = = + + Try This: Example 3A Solve: x 27 535 x + 27 = 535 –27 – 27
A. Isabelle earned $27 interest and now has a balance of $535 in the bank. What was her balance before interest was added?
balance before interest
interest earned + =
balance after interest + =
Solve: x 27
535
x + 27 = 535
–27
– 27
Subtract 27 from both sides.
x + 0 = 508
x = 508
Isabelle had a balance of $508 before interest was added.

25. = = + + Try This: Example 3B Solve: 472 n 502 472 + n = 502 –472 – 472
B. From June to July, the water level in a lake has increased from 472 feet to 502 feet. What was the increase in water level during that 1-month period?
initial water level
increase in water level + =
water level after increase + =
Solve:
472 n
502
472 + n = 502
–472
– 472
Subtract 472 from both sides.
0 + n = 30
n = 30
The increase in water level was 30 feet.

26. Lesson Quiz
Determine which value of x is a solution of the equation.
1. x + 9 = 17; x = 6, 8, or 26
2. x – 3 = 18; x = 15, 18, or 21
Solve.
3. a + 4 = 22
4. n – 6 = 39
5. The price of your favorite cereal is now $4.25. In prior weeks the price was $3.69. Write and solve an equation to find n, the increase in the price of the cereal. 8 21
a = 18
n = 45
n = 4.25; $0.56