I am attempting a basic affine cypher decryption and am left with the following systems of equations:
$$25a +b = 13 \mod 26 \text{ }(1)$$
$$14a +b = 6 \mod 26 \text{ }(2)$$
Is it correct to subtract $(2)$ from $(1)$ to get:
$$11a = 7 \mod 26$$
and solve the resulting equation for $a$? And then sub this value back into $(1)$ or $(2)$ and solve for $b$? Thanks!
Yes, what you suggested is fine.
Use Euclidean algorithm to find $11^{-1}\pmod{26}$.
$$26=2(11)+4$$
$$11=2(4)+3$$
$$4=3+1$$
Hence \begin{align}1&=4-(11-2(4)) \\ &=3(4)-11\\ &=3(26-2(11))-11 \\ &=3(26)-7(11)\end{align}
That is $$11^{-1}\equiv-7 \pmod{26}$$
$$a \equiv (-7)(7) \equiv -49 \equiv 52-49 \equiv 3 \pmod{26}$$
$$b \equiv 6-14(3) \equiv -36 \equiv -10 \equiv 16 \pmod{26}$$