Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.6 Rational Expressions and Equations - R.6 Exercise Set - Page 981: 47


The solution set of the rational equation $\frac{3}{y+7}=\frac{1}{y-8}$ is $\left\{ \frac{31}{2} \right\}$.

Work Step by Step

$\frac{3}{y+7}=\frac{1}{y-8}$ Substitute $y+7=0$ and $y-8=0$ Thus, $y=-7$ and $y=8$ The equation $\frac{3}{y+7}=\frac{1}{y-8}$ is undefined for $y=-7$and $y=8$, so $y\ne -7$and $y\ne 8$ Now, the denominators of the rational expressions are $y+7$, and $y-8$. On solving the LCD is $\left( y+7 \right)\left( y-8 \right)$. Multiply both sides by $\left( y+7 \right)\left( y-8 \right)$. $\left( y+7 \right)\left( y-8 \right)\left( \frac{3}{y+7} \right)=\left( y+7 \right)\left( y-8 \right)\left( \frac{1}{y-8} \right)$ Simplify the terms, $\begin{align} & \frac{3\left( y+7 \right)\left( y-8 \right)}{\left( y+7 \right)}=\frac{\left( y+7 \right)\left( y-8 \right)}{\left( y-8 \right)} \\ & 3\left( y-8 \right)=\left( y+7 \right) \end{align}$ Apply the distributive property, $\begin{align} & 3\left( y-8 \right)=\left( y+7 \right) \\ & 3y-24=y+7 \end{align}$ Apply the addition principle, subtract $y$ from both the sides of the equation: $\begin{align} & 3y-24-y=y+7-y \\ & 2y-24=7 \end{align}$ Again, apply the addition principle, add $24$ from both the sides of the equation: $\begin{align} & 2y-24+24=7+24 \\ & 2y=31 \end{align}$ Apply the multiplication principle: dividing $2$ from both the sides of the equation: $\begin{align} & \frac{2y}{2}=\frac{31}{2} \\ & y=\frac{31}{2} \end{align}$ Now, check the solution of the equation $\frac{3}{y+7}=\frac{1}{y-8}$, Substitute $y=\frac{31}{2}$ in $\frac{3}{y+7}=\frac{1}{y-8}$ $\begin{align} & \frac{3}{\frac{31}{2}+\frac{7\cdot 2}{2}}\overset{?}{\mathop{=}}\,\frac{1}{\frac{31}{2}-8} \\ & \text{ }\frac{3}{\frac{31}{2}+\frac{14}{2}}\overset{?}{\mathop{=}}\,\frac{1}{\frac{31}{2}-8} \\ & \text{ }\frac{3}{\frac{31+14}{2}}\overset{?}{\mathop{=}}\,\frac{1}{\frac{31}{2}-8} \\ \end{align}$ On further simplification: $\begin{align} & \frac{3\cdot 2}{31+14}\overset{?}{\mathop{=}}\,\frac{2}{31-16} \\ & \text{ }\frac{6}{45}\overset{?}{\mathop{=}}\,\frac{2}{15} \\ & \text{ }\frac{2}{15}=\frac{2}{15} \\ \end{align}$ This is true.
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