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hendricks_beginning_algebra_1e_ch1_3

114 Chapter 2 Linear Equations and Inequalities in One Variable 3 2 a = 12 1d. 2 3 a3 2 ab = 2 3 (12) Multiply each side by the reciprocal of 3 2 , 2 3 . 1a = 24 3 Simplify each side. a =8 Simplify. 3 2 a = 12 Original equation Check: 3 2 (8) = 12 Replace a with 8. 24 2 = 12 Simplify each side. 12 = 12 Simplify. Since a = 8 makes the equation true, the solution set is {8}. x 4 1e. =-8 1 4 x=-8 Rewrite as x 4 - 1 4 x. 4a1 4 xb = 4(-8) Multiply each side by the reciprocal of 1 4 , 4. x=-32 Simplify each side. x 4 Check: =-8 Original equation -32 4 =-8 Replace x with -32. -8=-8 Simplify. Since x = -32 makes the equation true, the solution set is {-32}. Student Check 1 Use the multiplication property of equality to solve each equation. a. 2x=-14 b. 0.4x = 24 c. -x=-8 d. 5 4 a=-20 e. y 3 =-15 Solving Linear Equations with the Addition and Multiplication Properties of Equality Most linear equations cannot be solved using only the addition property of equality or only the multiplication property of equality; they require the use of both properties. The addition property of equality is used to isolate the variable on one side of an equation and the multiplication property of equality is used to get a coefficient of one on the variable. Example 2 illustrates the use of both the addition property and multiplication property of equality to solve linear equations. �� ������������������������ ������������������ Solve each equation. 2a. 2x + 1 = -5 2b. 4y + 3 = -2y - 9 2c. 5x - 3 - 2x = 8x - 1 2d. 8(2x - 1) = 10x + 7 Objective 2 ▶ Apply both the addition and multiplication properties of equality to solve linear equations.


hendricks_beginning_algebra_1e_ch1_3
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