Answer
It is true.
The required solution set is $\left\{ 3,-1 \right\}$.
Work Step by Step
Consider the equations.
$\begin{align}
& 4x-3y=15 \\
& 3x+5y=4 \\
\end{align}$
$4x-3y=15$ …… (1)
$3x+5y=4$ …… (2)
Solve equation (2) for x.
$\begin{align}
& 3x+5y=4 \\
& 3x=4-5y \\
& x=\frac{4-5y}{3}
\end{align}$
Substitute $x=\frac{4-5y}{3}$ in equation (1) and solve for y.
$\begin{align}
& 4\left( \frac{4-5y}{3} \right)-3y=15 \\
& 16-20y-9y=45 \\
& 16-45=29y \\
& 29=29y
\end{align}$
Simplify further as follows.
$\begin{align}
& y=-\frac{29}{29} \\
& y=-1
\end{align}$
Substitute $y=-1$ in $x=\frac{4-5y}{3}$ and solve for x.
$\begin{align}
& x=\frac{4-5\left( -1 \right)}{3} \\
& =\frac{4+5}{3} \\
& =\frac{9}{3} \\
& =3
\end{align}$
Thus, $x=3,y=-1$.
Check:
Substitute $x=3,y=-1$ in equation (1).
$\begin{align}
4\left( 3 \right)-3\left( -1 \right)\overset{?}{\mathop{=}}\,15 & \\
12+3\overset{?}{\mathop{=}}\,15 & \\
15\overset{?}{\mathop{=}}\,15 & \\
\end{align}$
It is true.
