1. Lesson 1.3
Essential Question: How do I solve an equation with variables on both sides of an equal sign.
Objective: To solve equations with variables on both sides of the equal sign
Example:
12 + 4x = 10x
Subtract 4x from both sides.
– 4x – 4x
= 6x
Divide both sides by 6. 6 6
2 = x
Solution
Check: (2) = 10(2)
20 = 20

2. Example:
50 + 9x = 4x
Subtract 9x from both sides.
– 9x – 9x
50 = – 5x
Divide both sides by – 5.
– – 5
Solution
– 10 = x
Check: (-10) = 4(-10)
50 + (-90) = -40
– 40 = – 40

3. Example:
Distribute the
5x – 10 = 3 + 2x
Subtract 2x from both sides.
– 2x – 2x
Add 10 to both sides
3x – 10 = 3
3x = 13
Divide both sides by 3
x = or

4. Sometimes an equation does not have only one solution
Sometimes an equation does not have only one solution. Some equations have many solutions while others have no solutions.
Identity: Is an equation that is true for all values of the variable. (All real numbers are true)
Example:
4(x + 2) = 8 + 4x
Distribute the 4
4x + 8 = 8 + 4x
Subtract 4x from both sides
– 4x – 4x
True, Therefore this is an identity and all real numbers are solutions
8 = 8
All real numbers
To check, Replace x with any number

5. 4 = 7 4 = 7 Example: 9x + 4 = 7 + 9x Subtract 9x from both sides
False 4 does not equal 7 therefore there are no solutions. There is no number that will make this true.
4 = 7
No Solution

6. = + Verbal Models/Problem Solving Example:
Hank’s video store charges $8.00 to rent a video game for five days and does not charge an annual membership fee. Bunker’s video only charges $3.00 for a five day rental but has a $50.00 membership fee per year.
Find the number of rentals that would cost the same from each store.
Hanks
Number =
Bunkers
Number +
Membership
Charges
Rented
Charges
Rented
Fee
• x = • x
8x = 3x + 50
Subtract 3x from both sides
5x = 50
Divide both sides by 5
X = 10
If you rent 10 games per year, the cost would be the same.

7. Review your left column questions.
Suggestion: How do you tell the difference between an equation with no solutions and many solutions?
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