An algebra problem by A Former Brilliant Member

Given , $\dfrac{a - x}{b - x} = \dfrac{a^2}{b^2} \\ \implies (a - x)b^2 = (b - x)a^2 \\ \implies ab^2 - b^2x = a^2b - a^2x \\ \implies a^2x -b^2x = a^2b -ab^2 \\ \implies x(a^2 - b^2) = ab(a - b) \\ \implies x = \dfrac{ab(a - b)}{(a - b)(a + b)} \\ \therefore x = \boxed{\dfrac{ab}{a + b}}$

$\dfrac{a-x}{b-x}=\dfrac{a^2}{b^2}$ $\implies$ cross-multiplying, we get

$ab^2-xb^2=ba^2-xa^2$

then,

$xa^2-xb^2=ba^2-ab^2$

then,

$x(a^2-b^2)=ba^2-ab^2$ $\implies$ dividing both sides by $a^2-b^2$ , we get

$x=\dfrac{ba^2-ab^2}{a^2-b^2}$

then,

$x=\dfrac{ab(a-b)}{(a-b)(a+b)}=\boxed{\dfrac{ab}{a+b}}$

Given the equality in the problem, we have (a-x)b^2 = (b-x)a^2 Then, (a)(b^2) - (x)(b^2) = (b)(a^2) - (x)(a^2) Then we reorganize the equation to get (a)(b^2) - (b)(a^2) = (x)(b^2) - (x)(a^2) Factoring x in the right side and ab in the left side we have (ab)(b - a) = (x)(b^2 - a^2) Then we clear x by dividing so we get (ab)(b - a) / (b + a)(b - a) = x Simplify (b - a) and we finally get x = (ab) / (a+b)