1. Ch 2.5 Variable on Both Sides Objective: To solve equations where one variable exists on both sides of the equation.

2. GOAL: Isolate the variable on one side of the equation. 1) Use the Distributive Property. (then simplify by combining LIKE Terms) 2) Choose one of the variable expressions and use the Inverse Property of Addition 3) Apply the Inverse Property of Addition and/or the Inverse Property of Multiplication to the numbers. Perform Inverse operations to both sides of the equation! Rules

3. Special Cases 1)x = x 2) x + 1 = x Plug in various numbers for x …….. Solution: x = All Real Numbers Solution: x = No solution Every number makes a TRUE statement! Every number makes a FALSE statement! -x 0 = 0 -x 1 = 0

4. -3 - 3 2x + 4 = 5x - 17 -2x 4 = 3x - 17 +17 21 = 3x 3 7 = x -5x -3x + 4 = -17 -4 -3x = -21 x = 7 Example 1 Option 1: Subtract 2x from both sides Option 2: Subtract 5x from both sides

5. 4(x - 2) - 2x = 5(x - 4) 4x - 8 - 2x = 5x - 20 2x - 8 = 5x - 20 -2x -8 = 3x - 20 +20 12 = 3x 3 Example 2  Distributive Property  Combine LIKE Terms  Inverse Property of Addition for the variable  Inverse Property of Addition  Inverse Property of Multiplication

6. 3x + 8 = 2(x + 4) + x 3x + 2 = 2(x - 1) + x 3x + 8 = 2x + 8 + x 3x + 8 = 3x + 8 -3x 8 = 8 3x + 2 = 2x - 2 + x 3x + 2 = 3x - 2 -3x 2 = -2 x = any real number True ! False ! No Solution Example 3Example 4

7. 1) 3x - 5 = 2x + 12 -2x x - 5 = 12 +5 x = 17 2) 5x - 3 = 13 – 3x +3x 8x - 3 = 13 +3 8x = 16 8 x = 2 Classwork 3) 2b + 6 = 7b - 9 -2b 6 = 5b - 9 +9 15 = 5b 5 5 3 = b 4) -4c - 11 = 4c + 21 +4c -11 = 8c + 21 -21 -32 = 8c 8 8 -4 = c

8. 5) 3(x + 2) - (2x - 4) = - (4x + 5) 3x + 6 - 2x + 4 = - 4x - 5 x + 10 = - 4x - 5 + 4x 5x + 10 = -5 - 10 -10 5x = -15 5 x = -3

9. 6) 4(y - 2) + 6y = 7(y - 8) - 3(10 - y) 4y - 8 + 6y = 7y - 56 - 30 + 3y 10y - 8 = 10y - 86 -10y -8 = -86 False No Solution

10. 7) 3(4 + k) - 2(3k + 4) = 5(k - 3) - (8k - 19) 12 + 3k - 6k - 8 = 5k - 15 - 8k + 19 -3k + 4 = -3k + 4 +3k 4 = 4 True Infinitely Many Solutions! x = all real numbers

11. 8) 5(m - 4) = 10 - 4[2(m - 5) - 5m] 5m - 20 = 10 - 4[2m - 10 - 5m] 5m - 20 = 10 - 4[-3m - 10] 5m - 20 = 10 + 12m + 40 5m - 20 = 12m + 50 -5m -20 = 7m + 50 -50 -70 = 7m 7 x = -10