1. Solving Equations with Variables on Both Sides
10-3
Warm Up
Problem of the Day
Lesson Presentation
Warm-Up Exploration (7 minutes)
New homework points(3 minutes)
Homework check and questions(5 minutes)
Lesson 10.3 (15 minutes)
Finish Centers (10 minutes)
Pre-Algebra

7. What are some observations you made about solving equations with variables on both sides, based on last night’s homework?
Goal is to isolate the variable – get variables on one side and numbers on the other.
You can go in different order. There are different ways to go about the problem.

8. What can NOT do?
Ex A)
4x + 6 = x
4x + 6 = x
4x + 6 = x
– 4x – 4x
–x -6 – x -6
6 = –3x
3x = –6 3 6 –3
–3x =
–2 = x
x = -2

9. Ex B)
9b – 6 = 5b + 18
9b – 6 = 5b + 18
– 5b – 5b
4b – 6 = 18
4b = 24 4b 4 24 =
b = 6

10. With each step, you are keeping the equation, ajd just simplifying
With each step, you are keeping the equation, ajd just simplifying. If your final answer turns out to be not true, there is NO SOLUTION that would work for this equation.
Ex.C)
9w + 3 = 5w w
9w + 3 = 5w w
9w + 3 = 9w + 7
– 9w – 9w
3 ≠
No solution. There is no number that can be substituted for the variable w to make the equation true.

11. Ex. D y 5 3y 5 3 4 7 10
– = y –
20( ) = 20( ) y 5 3 4 3y 7 10
– y –
Multiply by the LCD.
20( ) + 20( ) – 20( )= 20(y) – 20( ) y 5 3y 3 4 7 10
4y + 12y – 15 = 20y – 14
16y – 15 = 20y – 14
Combine like terms.

12. 16y – 15 = 20y – 14
– 16y – 16y
Subtract 16y from both sides.
–15 = 4y – 14
Add 14 to both sides.
–1 = 4y –1 4 4y =
Divide both sides by 4. -1 4
= y

13. Ex.E)
12z – 12 – 4z = 6 – 2z + 32
12z – 12 – 4z = 6 – 2z + 32
8z – 12 = –2z + 38
+ 2z z
10z – 12 =
10z = 50
10z 10 =
z = 5

14. Additional Example 3: Consumer Application
Jamie spends the same amount of money each morning. On Sunday, he bought a newspaper for $1.25 and also bought two doughnuts. On Monday, he bought a newspaper for fifty cents and bought five doughnuts. On Tuesday, he spent the same amount of money and bought just doughnuts. How many doughnuts did he buy on Tuesday?

15. Additional Example 3 Continued
First solve for the price of one doughnut.
d = price of one doughnut.
d = d
– 2d – 2d
= d
– – 0.50
= d
0.75 3 3d =
The price of one doughnut is $0.25.
0.25 = d

16. Additional Example 3 Continued
Now find the amount of money Jamie spends each morning.
Choose one of the original expressions. d
(0.25) = 1.75
Jamie spends $1.75 each morning.
Find the number of doughnuts Jamie buys on Tuesday.
Let n represent the number of doughnuts.
0.25n = 1.75
0.25n
0.25
1.75 =
Divide both sides by 0.25.
n = 7; Jamie bought 7 doughnuts on Tuesday.