How to find solutions for $a$ and $b$ where $9 \equiv 4a+b \pmod {26} $ and $10 \equiv 19a+b \pmod {26}$? $$9 \equiv 4a+b \pmod {26}$$
$$10 \equiv 19a+b \pmod {26}$$
How can I solve the following system? 

You can write the system in matrix form:
$$
\pmatrix{4 & 1 \\ 19 & 1}
\pmatrix{a \\ b }
=
\pmatrix{9 \\ 10 }
$$
The determinant of the matrix is $-15$, which is invertible mod $26$, and so you can use Cramer's rule. In this $2\times 2$ case, it's easy:
$$
\pmatrix{4 & 1 \\ 19 & 1}^{-1}
=
-\frac{1}{15}
\pmatrix{1 & -1 \\ -19 & 4}
=
-7
\pmatrix{1 & -1 \\ -19 & 4}
\bmod 26
$$
and so
$$
\pmatrix{a \\ b }
=
-7
\pmatrix{1 & -1 \\ -19 & 4}
\pmatrix{9 \\ 10 }
=
\pmatrix{7 \\ 7 }
\bmod 26
$$
