Question

Solve : $\frac{3}{x+y}+\frac{2}{x-y}=2$ and $\frac{9}{x+y}-\frac{4}{x-y}=1$

• A. $x=\frac{2}{5};y=\frac{7}{2}$
• B. $x=\frac{5}{2};y=\frac{1}{2}$
• C. $x=\frac{6}{7};y=\frac{4}{3}$
• D. $x=\frac{7}{3};y=\frac{1}{2}$

Answer 1

The correct option is B $x=\frac{5}{2};y=\frac{1}{2}$

Let $\frac{1}{x+y}=X$ and $\frac{1}{x-y}=Y$
$3X+2Y=2$ ...(i)
$9X-4Y=1$ ...(ii)

On multiplying (i) by 3 and (ii) by 1 and subtracting, we get
$9X+6Y=6$
$\underset{–}{\underset{-}{9}X\underset{+}{-}4Y=\underset{-}{1}}$
$10Y=5$

$\therefore Y=\frac{1}{2}$

On putting $Y=\frac{1}{2}$ in (i), we get
$3X+2\times \frac{1}{2}=2\Rightarrow X=\frac{1}{3}$

$\therefore \frac{1}{x+y}=\frac{1}{3}$ and $\frac{1}{x-y}=\frac{1}{2}$

$\Rightarrow x+y=3$ ...(iii)
$x-y=2$ ...(iv)

On adding (iii) and (iv), we get
$2x=5\Rightarrow x=\frac{5}{2}$

On putting $x=\frac{5}{2}$ in (iii), we get

$\frac{5}{2}+y=3\Rightarrow y=3-\frac{5}{2}=\frac{1}{2}$

$\therefore x=\frac{5}{2},y=\frac{1}{2}$