Multiple choice

Solving the following equation. $\dfrac{x-a}{a^2}+\dfrac{y-b}{b^2}=\dfrac{1}{x-b}-\dfrac{1}{y-a}-\dfrac{1}{a-b}=0$. we get $x=\dfrac{a^2}{b}, \dfrac{a(2b-a)}{b}; y=\dfrac{b^2}{a}, \dfrac{b(2a-b)}{a}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The given solution x = a^2/b and y = b^2/a can be verified by substitution into the original equations. The equation structure (x-a)/a^2 + (y-b)/b^2 = 0 with 1/(x-b) - 1/(y-a) - 1/(a-b) = 0 yields the stated values, making the claim true.

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