Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.4 - Solving Linear Equations - Exercise Set: 46

Answer

conditional equation

Work Step by Step

Distribute $\frac{1}{3}$ on the left side and $\frac{1}{5}$ on the right side, then combine like terms to obtain: $\frac{1}{3}(6z) + \frac{1}{3}(12) = \frac{1}{5}(20z) +\frac{1}{5}(30) - 8 \\2z + 4 = 4z+6-8 \\2z + 4 = 4z - 2$ Subtract $2z$ and add $2$ on both sides of the equation to obtain: $4+2=4z-2z \\6 = 2z \\\frac{6}{2} = \frac{2z}{2} \\3 = z$ The given equation has one solution so it is a conditional equation.
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