Alternate solution approach to system of equations?

Question

Consider the pair of equations $$x_2 = \frac{(b-1)x_1 + 0.5}{b}$$ $$x_2 = \frac{(a-1)x_1 + 0.5}{a}$$ where $a,b>0$ and $a\neq b$.

The only solution is $x_1=x_2=0.5$, independent of $a$ and $b$. This can be obtained by writing the above system as $Ax=b$ and inverting.

Question: Is there a simpler solution approach than matrix inversion?

It's odd to me that the solution is independent of $a$ and $b$ so perhaps there's some significant simplification I'm missing.

Answer 1

Multiplying by $$ab$$ and we get $$a(b-1)x+\frac{1}{2}a=b(a-1)+\frac{1}{2}b$$ simplifying $$x(ab-a-ab+b)=\frac{1}{2}(b-a)$$ The rest is for you!

Answer 2

Here is an intuitive way of thinking about the problem. By symmetry, we can see that for every "non-trivial" case that we must have $a = b$. This means that the only way $a \neq b$ is if both equations are always true.

Treating the (without loss of generality) second equation as a polynomial in terms of $a$ this means that $x_1$ and $x_2$ must be equal and furthermore, the constant term must be 0 which forces both variables to be .5.