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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?

user348382
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    When you divided by $x-a-b$ you threw away the root $x=a+b$. Just like in $r(t-1)=t(2t-5)$, if we cancel the $t$ we are losing the root $t=0$. – André Nicolas Jul 06 '16 at 19:17
  • I figured this was the problem... Is my entire solution wrong then, or is there a way of keeping this solution intact? – user348382 Jul 06 '16 at 19:18
  • At the cancellation stage, say the following. This has the obvious root $x=a+b$. And if $x\ne a+b$, we can divide by $x-a-b$ and obtain $\dots$. – André Nicolas Jul 06 '16 at 19:20
  • so in the future, whenever I eliminate an expression with x should I keep track of it? – user348382 Jul 06 '16 at 19:21
  • Either keep track or just before cancelling note what happens if the cancelled object is equal to $0$. Keeping track is more unpleasant. – André Nicolas Jul 06 '16 at 19:23
  • @ guest.When you have $pq=pr$ you have equivalently $(p=0 $ or $ [ p\ne 0\land q=r].)$ Practice writing with $\iff$ between clauses as much as possible. This requires more writing, but if you start with $A$ and you have $A\iff B\iff C\iff D\iff E$ you will not miss anything. – DanielWainfleet Jul 06 '16 at 22:28

1 Answers1

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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 $t=0$.

André Nicolas
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