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

2. 1-3 Solving Addition and Subtraction Equations Warm Up
Course 3
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. 1-3 Solving Addition and Subtraction Equations Problem of the Day
Course 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. 1-3 Solving Addition and Subtraction Equations
Course 3
Learn to solve equations using addition and subtraction.

5. Vocabulary 1-3 equation solve solution inverse operation
Solving Addition and Subtraction Equations
Course 3
Vocabulary
equation
solve
solution
inverse operation
isolate the variable
Addition Property of Equality
Subtraction Property of Equality

6. 1-3 Solving Addition and Subtraction Equations
Course 3
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.
Course 3

7. 1-3 Solving Addition and Subtraction Equations
Course 3
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
1-3
Solving Addition and Subtraction Equations
Course 3
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
1-3
Solving Addition and Subtraction Equations
Course 3
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. 1-3 Solving Addition and Subtraction Equations
Course 3
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.
Course 3

11. 1-3 Solving Addition and Subtraction Equations
Course 3
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
Course 3

12. 1-3 Solving Addition and Subtraction Equations
Course 3
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
Course 3

13. 1-3 Solving Addition and Subtraction Equations
Course 3
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 ? 

14. 1-3 Solving Addition and Subtraction Equations
Course 3
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 ? 

15. 1-3 Solving Addition and Subtraction Equations
Course 3
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 ? 

16. = = + + 1-3 Solving Addition and Subtraction Equations
Course 3
Additional Example 3A
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.

17. = = + + 1-3 Solving Addition and Subtraction Equations
Course 3
Additional Example 3B
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.

18. 1-3 Solving Addition and Subtraction Equations Lesson Quiz 8 21 a = 18
Course 3
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