There is a smart way to solve this 2x2 system of linear equations?

Question

I was helping a high school student to study system of equations and he showed me this problem:

\begin{align}\frac{x-a}{b} + \frac{y-b}{a} &= x-a \\ \frac{x+y-b}{a}+\frac{y+b}{b} &= 3 . \end{align}

I suggested to him to solve for one of the variables (e.g.. $x$) and then substitute, but the problem becomes painfully hard, because of all the literals, and fractions. So, there is a intelligent way to solve this system of equations?

Answer 1

Rearrange the question into this: $$(x-a)\left(\frac1b-1\right)+(y-b)\frac1a=0\\ (x-a)\frac1a+(y-b)\left(\frac1a+\frac1b\right)=0$$ Then, provided the determinant of the $2\times2$ matrix is non-zero, there is exactly one solution.

Answer 2

Given the system

$\frac{x-a}{b} + \frac{y-b}{a}=x-a\\ \frac{x+y-b}{a}+\frac{y+b}{b}=3$

it is worth rearranging prior to solving, as

$$ax-a^2-axb+yb-b^2+a^2b=0\\ bx+by-b^2+ya+ab-3ab=0$$

Separating the variables and constants,

$$a(1-b)x+yb=a^2+b^2-a^2b\tag 1$$ $$bx+(a+b)y=b^2+2ab\tag 2$$

The "slopes" of the two equations are generally unrelated, and except for multiplying by similar quantities across both equations, we can't isolate variables:

$$(1)\cdot (a+b)+(2)\cdot (-b)\\ \to a(a+b)(1-b)x+b(a+b)y-b^2x-b(a+b)y=(a+b)(a^2+b^2-a^2b)-b^3-2ab^2\\ (a(a+b)(1-b)-b^2)x=a^3+a^2b-ab^2-a^3b-a^2b^2\\ (a^2+ab-a^2b-ab^2-b^2)x=a(a^2+ab-a^2b-ab^2-b^2)\\ x=a\tag 3$$

With this solution in hand, we immediately see that $y=b$.