Section 3.3 Multiplication and Division Properties of Equality 149 Check: 10 10 1. 10x 50 Original equation 10152 50 Substitute 5 for x. ✓ True Example 1 1x 5 b. 28 4p To obtain a coefficient of 1 for the p term, divide both sides by 4, because Simplify. The solution is 7 and checks in the original equation. 7 1p c. y 34 Note that –y is the same as 1 y. To isolate y, we need a coefficient of positive 1. To obtain a coefficient of 1 for the y term, divide both sides by 1. Simplify. The solution is 34 and checks in the original equation. 1y 34 1y 34 The multiplication property of equality indicates that multiplying both sides of an equation by the same nonzero quantity results in an equivalent equation. For example, consider the equation .The variable x is being divided by 2.To solve for x, we need to reverse this process. Therefore, we will multiply by 2. Multiply both sides by 2. The expression can be written as .This process is called regrouping factors. The x coefficient is 1, because equals 1. x 2 x 2 6 2 6 x 12 The solution is 12. 22 2 1 x 12 22 2 x x2 2 2 x 12 x2 6 y 34 1y 1 34 1 7 p 4 4 1. 28 4 4p 4 50 50 112 1y2 11234 Answers 1. 8 2. 9 3. 19 Applying the Division Property of Equality Solve the equations. a. 10x 50 b. 28 4p c. y 34 Solution: a. 10x 50 To obtain a coefficient of 1 for the x term, divide both sides by 10, because Simplify. x 5 The solution is 5. 10x 10 50 10 Skill Practice Solve the equations. 1. 4x 32 2. 18 2w 3. 19 m Avoiding Mistakes In Example 1(b), the operation between 4 and p is multiplication. We must divide by 4 (rather than 4) so that the resulting coefficient on p is positive 1. TIP: In Example 1(c), we could have also multiplied both sides by 1 to obtain a coefficient of 1 for x. y 34
miller_prealgebra_2e_ch1_3
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