We have:
[tex] A=\left[\begin{array}{cc}3&-5\\4&1\end{array}\right]\qquad X=\left[\begin{array}{c}a\\b\end{array}\right]\qquad C=\left[\begin{array}{c}2\\10\end{array}\right][/tex]
so:
[tex]A\cdot X=C\\\\\\
\left[\begin{array}{cc}3&-5\\4&1\end{array}\right]\cdot\left[\begin{array}{c}a\\b\end{array}\right]=\left[\begin{array}{c}2\\10\end{array}\right]\\\\\\
\left[\begin{array}{c}3a-5b\\4a+1b\end{array}\right]=\left[\begin{array}{c}2\\10\end{array}\right]\\\\\\
\boxed{\begin{cases}3a-5b=2\\4a+b=10\end{cases}}[/tex]
And the solution of the system of equations:
[tex]\begin{cases}3a-5b=2\\4a+b=10\quad|\cdot5\end{cases}\\\\\\
\begin{cases}3a-5b=2\\20a+5b=50\end{cases}\\---------(+)\\\\3a-5b+20a+5b=2+50\\\\3a+20a=52\\\\23a=52\quad|:23\\\\\boxed{a=\dfrac{52}{23}=2\dfrac{6}{23}}\\\\\\
4a+b=10\\\\b=10-4a\\\\b=10-4\cdot\dfrac{52}{23}\\\\\\b=\dfrac{230}{23}-\dfrac{208}{23}\\\\\\\boxed{b=\dfrac{22}{23}}\\\\\\
\boxed{\begin{cases}a=2\dfrac{6}{23}\\\\b=\dfrac{22}{23}\end{cases}}[/tex]
