Answer
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.
