Question

If $y(3x-y):x(4x+y)=5:12$, then find the ratio $(x^2+y^2):(x+y{)}^2$.

Answer 1

$\frac{y(3x-y)}{x(4x+y)}=\frac{5}{12}\Rightarrow 12y(3x-y)=5x(4x+y)\Rightarrow 36xy-12y^2=20x^2+5xy\Rightarrow 20x^2-31xy+12y^2=0\Rightarrow 20x^2-15xy-16xy+12y^2=0\Rightarrow 5x(4x-3y)-4y(4x-3y)=0\Rightarrow (5x-4y)(4x-3y)=0$
$5x-4y=0\Rightarrow x=\frac{4y}{5}$ ---(1)
or $4x-3y=0\Rightarrow x=\frac{3y}{4}$ ---(2)
Using(1)
$\frac{x^2+y^2}{(x+y{)}^2}=\frac{{\left(\frac{4y}{5}\right)}^2+y^2}{{(\frac{4y}{5}+y)}^2}=\frac{\frac{16y^2}{25}+y^2}{{\left(\frac{9y}{5}\right)}^2}=\frac{\frac{41y^2}{25}}{\frac{81y^2}{25}}=\frac{41}{81}$