Number systems and binary
This feels kinda mathy for EE, but here goes.

I hope the reductions make sense. They rewrite a two digit number like '35 base y' as \$ 3y^1+5y^0 \$. This generalizes as \$ digitn \times y^n+... \$.

There's an error in the last line of math for the solution when they convert the relation from 4x = 3y to x>=3y/4. Those statements aren't equivalent.

We see a digit "9" in the left-hand number which means it's at least in base 10, and given we know it's in base x-1, then x >= 11.

By the same logic: We see a digit "5" in the right-hand number which means it's at least in base 6, and given we know it's in base y, then y >= 6.

We also know that x and y are both integers, so x is divisible by 3 so y is an integer, and y is divisible by 4 so x is an integer.

We can then check y = 8 and solve for x finding that if y = 8 then x = 6 which isn't high enough, so instead we pick the smallest value for x that is greater than 11 and divisible by 3. Then we solve for y.