<s>[INST] Why is my solution wrong? Question: "Find all x such that ${\frac{x-a}{b}}+{\frac{x-b}{a}}={\frac{b}{x-a}}+{\frac{a}{x-b}}$ , where a and b are constants."

My attempt:
${\frac{x-a}{b}}+{\frac{x-b}{a}}={\frac{b}{x-a}}+{\frac{a}{x-b}}$
Let $m=x-a$ ,Let $w=x-b$ . Therefore:
${\frac{m}{b}}+{\frac{w}{a}}={\frac{b}{m}}+{\frac{a}{w}}$
${\frac{m}{b}}-{\frac{b}{m}}={\frac{a}{w}}-{\frac{w}{a}}$
${\frac{m^2-b^2}{bm}}={\frac{a^2-w^2}{aw}}$
${\frac{(m+b)(m-b)}{bm}}={\frac{(a+w)(a-w)}{aw}}$
Substituting back $m=x-a$ , $w=x-b$ :
${\frac{(x-a+b)(x-a-b)}{b(x-a)}}={\frac{(a+x-b)(a-x+b)}{a(x-b)}}$
Factoring out $-1$ from $(a-x+b)$:
${\frac{(x-a+b)(x-a-b)}{b(x-a)}}={\frac{-(a+x-b)(x-a-b)}{a(x-b)}}$
Dividing by $(x-a-b)$:
${\frac{(x-a+b)}{b(x-a)}}={\frac{-(a+x-b)}{a(x-b)}}$
Cross multiplying:
$(x-a+b)(ax-ab)=(-x-a+b)(bx-ab)$
Distributing;
$ax^2-abx-a^2x+a^2b+abx-ab^2=-bx^2+abx-abx+a^2b+b^2x-ab^2$
$ax^2-a^2x+a^2b-ab^2=-bx^2+a^2b+b^2x-ab^2$
$ax^2-a^2x+a^2b-ab^2-(-bx^2+a^2b+b^2x-ab^2)=0$
$ax^2+bx^2-a^2x-b^2x=0$
$(a+b)x^2-(a^2+b^2)x=0$
Factoring out $x$:
$x[(a+b)x-(a^2+b^2)]=0$
Therefore:
$x=0$
or
$(a+b)x-(a^2+b^2)=0$
$(a+b)x=(a^2+b^2)$
$x=\frac{(a^2+b^2)}{(a+b)}$

My two solutions are $x=0$ and $x=\frac{(a^2+b^2)}{(a+b)}$; however, I am missing the solution $x=a+b$. Where did I eliminate this solution? How do I prevent this from happening (eliminating solutions) in the future? Is there a better way to solve this equation?
 [/INST] When you divided by $x-a-b$ you threw away the root $x=a+b$.  Just like in the equation $t(t-1)=t(2t-5)$, if we cancel the $t$ we are losing the root $t=0$.
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