How to find solutions for $a$ and $b$ where $9 \equiv 4a+b \pmod {26} $ and $10 \equiv 19a+b \pmod {26}$?

Question

$$9 \equiv 4a+b \pmod {26}$$

$$10 \equiv 19a+b \pmod {26}$$

How can I solve the following system?

Answer 1

Notice that substracting the two equations gives us that $15a \equiv 1 \pmod {26}$.

This gives us that $a \equiv 7 \pmod {26}$.

Since $4 \times 7+b \equiv 9 \equiv 35 \equiv 5 \times 7 \pmod {26}$, this implies that $b \equiv 7 \pmod {26}$.

The answer is $a \equiv b \equiv 7 \pmod {26}$.

Answer 2

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 $$ \pmatrix{a \\ b } = -7 \pmatrix{1 & -1 \\ -19 & 4} \pmatrix{9 \\ 10 } = \pmatrix{7 \\ 7 } \bmod 26 $$