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.