Solving Equations With Variables on Both Sides Objective: Solve equations with the variable on each side and solve equations involving grouping symbols.

Variables on Each Side To solve an equation that has variables on each side, use the Addition or Subtraction Property of Equality to write an equivalent equation with the variable terms on one side. Collect the variable terms on the side of the equation with the larger variable coefficient (number in front of variable). Collect the constant terms (numbers) on the other side of the equation. Solve as a multi-step equation in one variable.

Example 1 Solve 8 + 5c = 7c – 2. Check your solution c = 7c – 2 -5c 8 = 2c – = 2c 2 5 = c c = c = 7c – (5) = 7(5) – = 35 – 2 33 = 33 

Grouping Symbols If equations contain grouping symbols such as parenthesis or brackets, use the Distributive Property first to remove the grouping symbols.

Example 2 Solve the following equations. Check your solution. a. 1 / 3 ( q) = 6(2q – 7) 1 / 3 (18) + 1 / 3 (12q) = 6(2q) – 6(7) 6 + 4q = 12q – 42 -4q 6 = 8q – = 8q 8 6 = q q = 6 1 / 3 ( q) = 6(2q – 7) 1 / 3 ( ) = 6(2 6 – 7) 1 / 3 ( ) = 6(12 – 7) 1 / 3 (90) = 6(5) 30 = 30 

Example 2 Solve the following equations. Check your solution. b x + 2 = 9x -5x 2 = 4x 4 ½ = x x = ½ = 

Special Solutions Some equations may have no solution. That is, there is no value of the variable that will result in a true equation. Some equations are true for all values of the variables. These are called identities.

Example 3 Solve each equation. a.8(5c – 2) = 10(32 + 4c) 8(5c) – 8(2) = 10(32) + 10(4c) 40c – 16 = c -40c -16 = 320 x No Solution

Example 3 Solve each equation. b.4(t + 20) = 1 / 5 (20t + 400) 4(t) + 4(20) = 1 / 5 (20t) + 1 / 5 (400) 4t + 80 = 4t t 80 = 80 Identity (True for all values of t)

Steps for Solving Equations 1.Simplify the expressions on each side. Use the Distributive Property as needed. 2.Use the Addition and/or Subtraction Properties of Equality to get the variables on one side and the numbers without variables on the other side. Simplify. 3.Use the Multiplication or Division Property of Equality to solve.

Example 4 Find the value of h so that the figures have the same area. A.1 B.3 C.4 D.5 h 6 (h – 2) 10 ½ (6)h = ½ (10)(h – 2) 3h = 5(h – 2) 3h = 5(h) – 5(2) 3h = 5h – 10 -3h 0 = 2h – = 2h 2

Check Your Progress Choose the best answer for the following. Solve 9f – 6 = 3f + 7. A. 1 / 3 B. 1 / 12 C.2 1 / 6 D.2 9f – 6 = 3f f 6f – 6 = f = 13 6

Check Your Progress Choose the best answer for the following. Solve 6(3r – 4) = 3 / 8 (46r + 8). A.38 B.28 C.10 D.36 6(3r) – 6(4) = 3 / 8 (46r) + 3 / 8 (8) 18r – 24 = 69 / 4 r r -24 = - ¾ r = - ¾ r - 4 / 3

Check Your Progress Choose the best answer for the following. Solve 2(4a + 8) = 3( 8a / 3 – 10). A. 1 / 3 B.2 C.True for all values of a. D.No solution. 2(4a) + 2(8) = 3( 8a / 3 ) – 3(10) 8a + 16 = 8a – 30 -8a 16 = -30

Check Your Progress Choose the best answer for the following. Solve 1 / 7 (21c – 56) = 3(c – 8 / 3 ). A. 1 / 3 B.0 C.True for all values of c. D.No solution. 1 / 7 (21c) – 1 / 7 (56) = 3(c) – 3( 8 / 3 ) 3c – 8 = 3c – 8 -3c -8 = -8

Check Your Progress Choose the best answer for the following. Find the value of x so that the figures have the same area. A.1 B.2 C.3 D.4 6 x 4 x + 4 6x = ½ (4)(x + 4) 6x = 2(x + 4) 6x = 2(x) + 2(4) 6x = 2x x 4x = 8 4