Lesson 1 3 Essential Question How do I

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 + 4 x = 10 x – 4 x 12 = 6 x 6 6 2=x Subtract 4 x from both sides. Divide both sides by 6. Solution Check: 12 + 4(2) = 10(2) 20 = 20

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

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

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 + 4 x 4 x + 8 = 8 + 4 x – 4 x 8=8 All real numbers Distribute the 4 Subtract 4 x from both sides True, Therefore this is an identity and all real numbers are solutions To check, Replace x with any number

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

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 Rented Charges 8 • x = Bunkers Charges = 3 Number Rented • x Membership + Fee + 50 Subtract 3 x from both sides 8 x = 3 x + 50 5 x = 50 Divide both sides by 5 X = 10 If you rent 10 games per year, the cost would be the same.

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