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 - Page 363: 90


The solution set is\[\left\{ -\frac{5}{2} \right\}\].

Work Step by Step

The equation is\[\frac{x+4}{8}=\frac{3}{16}\]. Apply the cross-product principle in the equation. \[\begin{align} & \frac{x+4}{8}=\frac{3}{16} \\ & 16\left( x+4 \right)=3\times 8 \\ & 16x+64=18 \\ \end{align}\] Subtract \[64\]from both sides of the equal sign. \[\begin{align} & 16x+64-64=18-64 \\ & 16x=-40 \\ \end{align}\] Divided by \[16\], both sides of the equal sign. \[\begin{align} & \frac{16x}{16}=\frac{-40}{16} \\ & x=-\frac{5}{2} \\ \end{align}\] Check the proposed solution. Substitute \[\frac{-5}{2}\] for x in the original equation\[\frac{x+4}{8}=\frac{3}{16}\]. \[\begin{align} & \frac{\frac{-5}{2}+4}{8}=\frac{3}{16} \\ & \frac{\frac{-5+8}{2}}{8}=\frac{3}{16} \\ & \frac{3}{2\times 8}=\frac{3}{16} \\ & \frac{3}{16}=\frac{3}{16} \\ \end{align}\] This true statement \[\frac{3}{16}=\frac{3}{16}\] verifies that the solution set is\[\left\{ -\frac{5}{2} \right\}\]. Thus, the solution set is\[\left\{ -\frac{5}{2} \right\}\].
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