Question

Root ax-root by=b-a and root bx-root ay=0,then x+ y is

Answer 1

$\text{Given equations are}$

$\sqrt{ax}-\sqrt{by}=b-a$

$\sqrt{bx}-\sqrt{ay}=0$

$\text{It can be written as}$

$\sqrt{a}\sqrt{x}-\sqrt{b}\sqrt{y}-(b-a)=0$

$\sqrt{b}\sqrt{x}-\sqrt{a}\sqrt{y}=0$

$\text{By cross multiplication rule, we get}$

$\displaystyle\frac{\sqrt{x}}{0-(b-a)\sqrt{a}}=\frac{\sqrt{y}}{-(b-a)\sqrt{b}-0}=\frac{1}{a+b}$

$\displaystyle\frac{\sqrt{x}}{(a-b)\sqrt{a}}=\frac{\sqrt{y}}{(a-b)\sqrt{b}}=\frac{1}{a+b}$

$\implies\displaystyle\frac{\sqrt{x}}{(a-b)\sqrt{a}}=\frac{1}{a+b}$

$\implies\displaystyle\;\sqrt{x}=\frac{(a-b)\sqrt{a}}{a+b}$

$\implies\bf\displaystyle\;x=\frac{(a-b)^2a}{(a+b)^2}$

$\text{Similarly,}$

$\implies\bf\displaystyle\;y=\frac{(a-b)^2b}{(a+b)^2}$

$\text{Now,}$

$x+y$

$=\displaystyle\frac{(a-b)^2a}{(a+b)^2}+\frac{(a-b)^2b}{(a+b)^2}$

$=\displaystyle\frac{(a-b)^2(a+b)}{(a+b)^2}$

$=\displaystyle\frac{(a-b)^2}{a+b}$

$\implies\boxed{\bf\;x+y=\frac{(a-b)^2}{a+b}}$