Answer
$(\displaystyle \frac{4}{3},\frac{1}{5})$
Work Step by Step
Cramer's rule
$\left\{\begin{array}{l}{a x+b y=s}\\{cx+dy=t}\end{array}\right.$
$D=\left|\begin{array}{ll}{a}&{b}\\{c}&{d}\end{array}\right|,D_{x}=\left|\begin{array}{ll}{s}&{b}\\{t}&{d}\end{array}\right|,D_{y}=\left|\begin{array}{ll}{a}&{s}\\{c}&{t}\end{array}\right|,$
If $D\displaystyle \neq 0,\qquad x=\frac{D_{x}}{D}\quad y=\frac{D_{y}}{D}$
---
$\left\{\begin{aligned}3x-5y&=3\\15x+5y&=21\end{aligned}\right.\Rightarrow\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
3 & -5\\
15 & 5
\end{array}\right],\quad \left[\begin{array}{l}
s\\
t
\end{array}\right]=\left[\begin{array}{l}
3\\
21
\end{array}\right]$
$\begin{array}{lllll}
D=\left|\begin{array}{ll}
3 & -5\\
15 & 5
\end{array}\right|= & & D_{x}=\left|\begin{array}{ll}
3 & -5\\
21 & 5
\end{array}\right|= & & D_{y}=\left|\begin{array}{ll}
3 & 3\\
15 & 21
\end{array}\right|=\\
=15+75 & & =15+105 & & =63-45\\
=90\neq 0 & & =120 & & =18\\
& & & &
\end{array}$
$x=\displaystyle \frac{D_{x}}{D}=\frac{120}{90}=\frac{4}{3}$
$y=\displaystyle \frac{D_{y}}{D}=\frac{18}{90}=\frac{1}{5}$
Solution: $(x,y)=(\displaystyle \frac{4}{3},\frac{1}{5})$
