<s>[INST] 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? 
 [/INST] 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 $a \equiv b \equiv 7 \pmod {26}$. 
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