Question

In the system of equations $\frac{6}{x+y}-\frac{3}{x-y}=2$ and $\frac{4}{x+y}+\frac{6}{x-y}=4$, the value of x and y will be

• A. 3 and $-\frac{1}{3}$
• B. $\frac{5}{2}$ and $-\frac{1}{2}$
• C. $\frac{4}{3}$ and $-\frac{2}{3}$
• D. $\frac{5}{2}$ and $\frac{1}{4}$

Answer 1

The correct option is B $\frac{5}{2}$ and $-\frac{1}{2}$

$\because \frac{6}{x+y}-\frac{3}{x-y}=2$ ......(i)

and $\frac{4}{x+y}+\frac{6}{x-y}=4$ ......(ii)

If $\frac{1}{x+y}=a$ and $\frac{1}{x-y}=b$, then the equation will be:

$\Rightarrow$$6a-3b=2$ .....(iii)

$\Rightarrow$$4a+6b=4$ .....(iv)

On multiplying equation (iii) by 2 and adding to equation (iv),

$\Rightarrow$$2(6a-3b)+4a+6b=2\times 2+4$

$\therefore 12a-6b+4a+6b=4+4$

$\Rightarrow 16a=8\Rightarrow a=\frac{8}{16}=\frac{1}{2}$

On substituting this value of a in equation (iii),

$6\times \frac{1}{2}-3b=2\Rightarrow 3b=1\Rightarrow b=\frac{1}{3}$

Since $\frac{1}{x+y}=a=\frac{1}{2}\Rightarrow x+y=2$ ......(v)

and $\frac{1}{x-y}=b=\frac{1}{3}\Rightarrow x-y=3$ .....(vi)

On adding equations (v) and (vi),

$2x=5\Rightarrow x=\frac{5}{2}$

On substituting this value of x in equation (v),

$\Rightarrow$$\frac{5}{2}+y=2$

$\therefore x=\frac{5}{2}$ and $y=-\frac{1}{2}$.