Question

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

Answer 1

$y(3x-y):x(4x+y)=5:12$

$\Rightarrow \frac{y(3x-y)}{x(4x+y)}=\frac{5}{12}$

$12y(3x-y)=5x(4x+y)$

$\Rightarrow 36xy-12y^2=20x^2+5xy$

$36xy-12y^2-20x^2-5xy=20x^2+5xy-20x^2-5xy$

$\Rightarrow -20x^2+31xy-12y^2=0$

$\Rightarrow 20x^2-16xy-15xy+12y^2=0$

$\Rightarrow 4x(5x-4y)-3y(5x-4y)=0$

$\Rightarrow (4x-3y)(5x-4y)=0$

Then if $4x-3y=0\Rightarrow 4x=3y\Rightarrow x=\frac{3}{4}y$.................(1)

And $5x-4y=0\Rightarrow 5x=4y\Rightarrow x=\frac{4}{5}y$.......................(2)

Then If $x=\frac{3}{4}y$

$\therefore x^2+y^2={\left(\frac{3}{4}y\right)}^2+y^2=\frac{9y^2}{16}+y^2=\frac{9y^2+16y^2}{16}=\frac{25y^2}{16}$.......................(4)

$\therefore (x+y{)}^2={(\frac{3}{4}y+y)}^2={\left(\frac{3y+4y}{4}\right)}^2={\left(\frac{7y}{4}\right)}^2=\frac{49y^2}{16}$..................(5)

If $x=\frac{5}{4}y$

$\therefore x^2+y^2={\left(\frac{5}{4}y\right)}^2+y^2=\frac{25}{16}{y}^2+y^2=\frac{25y^2+16y^2}{16}=\frac{41}{16}{y}^2$.................(6)

$\therefore (x+y{)}^2={(\frac{5}{4}y+y)}^2={\left(\frac{5y+4y}{4}\right)}^2={\left(\frac{9}{4}y\right)}^2=\frac{81}{16}{y}^2$.....................(7)

As per (4) and (5)

$x^2+y^2:(x+y{)}^2=\frac{25y^2}{16}:\frac{49y^2}{16}=25:49$

As per (6) and (7)

$x^2+y^2:(x+y{)}^2=\frac{41y^2}{16}:\frac{81y^2}{16}=41:81$