Answer
The solution set of the provided equation $\frac{x+3}{4}=\frac{x-2}{3}+\frac{1}{4}$ is $\left\{ 14 \right\}$.
Work Step by Step
Consider the provided equation is,
$\frac{x+3}{4}=\frac{x-2}{3}+\frac{1}{4}$
Take common denominator of the above equation and simplify,
Thus,
$\begin{align}
& \frac{x+3}{4}=\frac{x-2}{3}+\frac{1}{4} \\
& \frac{x+3}{4}=\frac{4\left( x-2 \right)+1\left( 3 \right)}{12} \\
& \frac{x+3}{4}=\frac{4x-8+3}{12} \\
& \frac{x+3}{4}=\frac{4x-5}{12}
\end{align}$
Cross multiply and add like terms of the above equation,
Thus,
$\begin{align}
& \frac{x+3}{4}=\frac{4x-5}{12} \\
& 12\left( x+3 \right)=4\left( 4x-5 \right) \\
& 12x+36=16x-20 \\
& 56=4x
\end{align}$
Divide both sides of the above equation by $4$ ,
Thus,
$\begin{align}
& \frac{4x}{4}=\frac{56}{4} \\
& x=14
\end{align}$
Hence, the solution set of the provided equation $\frac{x+3}{4}=\frac{x-2}{3}+\frac{1}{4}$ is $\left\{ 14 \right\}$.
