Question

If $\frac{4x^2+3y^2}{4x^2-3y^2}=\frac{11}{5}$, use properties of proportion to prove the $x^2=2y^2$.

Answer 1

$\frac{4x^2+3y^2}{4x^2-3y^2}=\frac{11}{5}$

Using compounded (Divided )

$\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a+b}{a-b}=\frac{c+d}{c-d}$

$\Rightarrow \frac{4x^2+3y^2+4x^2-3y^2}{4x^2+3y^2-4x^2+3y^2}=\frac{11+5}{11-5}$

$\Rightarrow \frac{8x^2}{6y^2}=\frac{16}{6}$

$\Rightarrow 8x^2\times 6=6y^2\times {16}^2$

$\Rightarrow x^2=2y^2$

Hence proved.