abstract algebra - How to solve system of equations with mod?

I'm trying to solve for $a$ and $b$:
$$5 \equiv (4a + b)\bmod{26}\quad\text{and}\quad22\equiv (7a + b)\bmod{26}.$$
I tried looking it up online, and the thing that seemed most similar was the Chinese remainder theorem; however, I couldn't find an instance where it fit something more like what I want to solve. A simple explanation or a reference to one would be most appreciated.

With my (limited) knowledge of algebra, I figured out that $x\ \textrm{mod}\ 26 = x - 26\lfloor\frac{x}{26}\rfloor$, so I tried substituting that into my equations:
$$5=(4a+b)-26\left\lfloor\frac{4a+b}{26}\right\rfloor\quad\text{and}\quad 22=(7a+b)-26\left\lfloor\frac{7a+b}{26}\right\rfloor.$$

And I figured I could do something with that since I got rid of the mod, but... I have never solved an equation with a floor function before.

Answer

Well, mod is easier to handle. We have only $m$ numbers $\pmod m$: $0,1,\dots,m-1$ and already $m\equiv 0$ (also, $-1\equiv m-1$), it goes in a cycle just like the hours in a day $\pmod{12}$.

Precisely, $a\equiv b \pmod m$ means $\ m|(a-b)$, and the arithmetic operations such as $+,-,\cdot$ are very friendly with it, $\equiv$ acts just like an equation.

You can try to solve it, like $b\equiv 22-7a \pmod{26}$, then substitute it back to the other, $5\equiv -3a+22$, so $3a\equiv 17$, but $\pmod{26}$ this $17$ can be substituted by $-9$ for example (because $17\equiv -9 \pmod{26}$), and $3$ is coprime to $26$ so one can divide by $3$.