Question

If $\frac{3x+2y}{3x-2y}=\frac{4}{3}$ then find the value of $\frac{x^2+y^2}{x^2-y^2}$

• A. $\frac{185}{205}$
• B. $\frac{205}{187}$
• C. $\frac{14}{3}$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{205}{187}$

Given,

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

Apply dividendo and componendo

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

$\frac{6x}{4y}=\frac{7}{1}$

$\frac{x}{y}=\frac{14}{3}-\left(1\right)$

Therefore

$\frac{x^2+y^2}{x^2-y^2}=\frac{y^2(\frac{x^2}{y^2}-1)}{y^2(\frac{x^2}{y^2}-1)}$

$=\frac{(\frac{14}{3}{)}^2+1}{(\frac{14}{3}{)}^2-1}$

$=\frac{196+9}{196-9}=\frac{205}{187}$