Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 6 - Algebra: Equations and Inequalities - 6.2 Linear Equations in One Variable and Proportions - Exercise Set 6.2: 44

Answer

z = -10

Work Step by Step

The equation as written is: -2(z - 4) - (3z - 2) = -2 - (6z - 2) We first need to apply the distributive property. -2z + 8 - 3z + 2 = -2 - 6z + 2 Now we can add/subtract like terms on each side of the equation. We can regroup the terms first. The sign to the left of a term stays with the term. -2z - 3z + 8 + 2 = -6z - 2 + 2 -5z + 10 = -6z Now, we can add 6z to both sides of the equation. -5z + 6z + 10 = -6z + 6z Completing the arithmetic: z + 10 = 0 Next, subtract 10 from both sides of the equations. z + 10 - 10 = 0 - 10 z = -10 (our solution) Checking the answer, we can substitute z = -10 back into the original equation. Once we do this and simplify both sides, we can check to see if the end result matches. If it does, our solution is correct. -2(z - 4) - (3z - 2) = -2 - (6z - 2) Substituting -10 for z, we get: -2(-10 - 4) - [(3)(-10) - 2] = -2 - [(6)(-10) - 2] Complete operations inside grouping symbols (if more than one operation is inside grouping, use order of operation --it may take more than one step) -2(-14) - (-30 - 2) = -2 -(-60 - 2) Keep working in grouping. -2(-14) - (-32) = -2 - (-62) Now, multiply left to right 28 + 32 = -2 + 62 Now we add.subtract on each side of the equation. 60 = 60 The answers match, so our solution of z = -10 is correct.
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