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: 43


The solution set of the rational equation $\frac{1}{2}+\frac{1}{3}=\frac{1}{t}$ is $\left\{ \frac{6}{5} \right\}$.

Work Step by Step

$\frac{1}{2}+\frac{1}{3}=\frac{1}{t}$ Note that, $t\ne 0$. The denominators of the rational expressions are $2$, $3$, and $t$. Multiply both sides by $6t$. $6t\left( \frac{1}{2}+\frac{1}{3} \right)=6t\left( \frac{1}{t} \right)$ Apply the distributive property, $6t\cdot \left( \frac{1}{2} \right)+6t\cdot \left( \frac{1}{3} \right)=6t\left( \frac{1}{t} \right)$ Simplify the terms, $\begin{align} & \frac{6t}{2}+\frac{6t}{3}=\frac{6t}{t} \\ & \frac{2\cdot 3\cdot t}{2}+\frac{2\cdot 3\cdot t}{3}=\frac{6t}{t} \\ & 3t+2t=6 \end{align}$ Again, apply the distributive property, $\begin{align} & 3t+2t=6 \\ & 5t=6 \end{align}$ Apply the Multiplication principle, divide both sides by 5: $\begin{align} & \frac{5t}{5}=\frac{6}{5} \\ & t=\frac{6}{5} \end{align}$ Now, check the solution of the equation$\frac{1}{2}+\frac{1}{3}=\frac{1}{t}$ Substitute $t=\frac{6}{5}$ in $\frac{1}{2}+\frac{1}{3}=\frac{1}{t}$: $\begin{align} & \frac{1}{2}+\frac{1}{3}\overset{?}{\mathop{=}}\,\frac{1}{\left( \frac{6}{5} \right)} \\ & \frac{1}{2}+\frac{1}{3}\overset{?}{\mathop{=}}\,\frac{5}{6} \\ & \frac{3+2}{6}\overset{?}{\mathop{=}}\,\frac{5}{6} \\ & \text{ }\frac{5}{6}=\frac{5}{6} \\ \end{align}$ Both sides of above equation are same. So, $t=\frac{6}{5}$ is the solution of $\frac{1}{2}+\frac{1}{3}=\frac{1}{t}$.
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