Answer
The solution set of the rational equation $\frac{x-7}{x+1}=\frac{2}{3}$ is $\left\{ 23 \right\}$.
Work Step by Step
$\frac{x-7}{x+1}=\frac{2}{3}$
Substitute, $x+1=0$
$\begin{align}
& x+1-1=0-1 \\
& x=-1
\end{align}$
Thus, the given equation $\frac{x-7}{x+1}=\frac{2}{3}$is undefined for $x=-1$. So $x\ne -1$
The denominators of the rational expressions are $x+1$, and 3.
On solving the LCD is $3\left( x+1 \right)$.
Multiply both sides by $3\left( x+1 \right)$.
$3\left( x+1 \right)\left( \frac{x-7}{x+1} \right)=3\left( x+1 \right)\left( \frac{2}{3} \right)$
Simplify the terms,
$\begin{align}
& \frac{3\left( x+1 \right)\left( x-7 \right)}{\left( x+1 \right)}=\frac{3\left( x+1 \right)2}{3} \\
& 3\left( x-7 \right)=2\left( x+1 \right)
\end{align}$
Apply the distributive property:
$3x-21=2x+2$
Apply the addition principle, subtracting $2x$ from both the sides of the equation:
$\begin{align}
& 3x-21-2x=2x+2-2x \\
& x-21=2
\end{align}$
Again, apply the addition principle, add $21$ on both the sides of the equation:
$\begin{align}
& x-21+21=2+21 \\
& x=23
\end{align}$
Now, check the solution of the equation $\frac{x-7}{x+1}=\frac{2}{3}$,
Substitute $x=23$ in $\frac{x-7}{x+1}=\frac{2}{3}$:
$\begin{align}
\frac{23-7}{23+1}\overset{?}{\mathop{=}}\,\frac{2}{3} & \\
\frac{16}{24}\overset{?}{\mathop{=}}\,\frac{2}{3} & \\
\frac{2}{3}=\frac{2}{3} & \\
\end{align}$
So, $x=23$ is the solution of $\frac{x-7}{x+1}=\frac{2}{3}$.
