Section 2.3 The Multiplication Property of Equality 113 Example 1 illustrates how we solve an equation that requires us to use only the multiplication property of equality. In the next objective and in Section 2.4, we will solve equations that require both the addition and multiplication properties of equality. ���������������������� Using the Multiplication Property of Equality to Solve Equations of the Form ax = b Step 1: Determine what operation will isolate the variable on one side of the equation. Remember the inverse property for multiplication: a · 1 a = a a = 1, that is, the product of reciprocals is 1. Step 2: Perform this operation on each side of the equation. Step 3: Simplify each side of the equation, as necessary. The result should be of the form x = some number or some number = x Step 4: Check the solution by substituting the value into the original equation. Step 5: Write the solution in set notation. �� ������������������������ ������������������ Use the multiplication property of equality to solve each equation. 1a. 3x=-12 1b. 0.05y = 40 1c. -x = 6 1d. 3 2 a = 12 1e. x 4 =-8 Solutions 1a. 3x=-12 3x 3 = -12 3 Divide each side by 3. x=-4 Simplify. Recall 3x 3 = 3 3 x = 1x = x. Check: 3x=-12 Original equation 3(-4)=-12 Replace x with -4. -12=-12 Simplify. Since x = -4 makes the equation true, the solution set is {-4}. 1b. 0.05y = 4 0.05y 0.05 = 4 0.05 Divide each side by 0.05. y = 80 Simplify. Recall 0.05y 0.05 = 0.05 0.05 y = 1y = y. Check: 0.05y = 4 Original equation 0.05(80) = 4 Replace y with 80. 4 = 4 Simplify. - Since y = 80 makes the equation true, the solution set is {80}. 1c. x = 6 -1x = 6 Rewrite -x as -1x. -1x 6 = -1 -1 Divide each side by -1. x=-6 Simplify. Check: -x = 6 Original equation -(-6) = 6 Replace x with -6. 6 = 6 Simplify. Since x = -6 makes the equation true, the solution set is {-6}.
hendricks_beginning_algebra_1e_ch1_3
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