Question

In the system of equations $\frac{12}{x+y}+\frac{8}{x-y}=8$ and $\frac{27}{x+y}-\frac{12}{x-y}=3$, the values of $x$ and $y$ will be

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

Answer 1

Answer: (B)

$\frac{5}{2}$ and $\frac{1}{2}$

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

$12a+8b=8$........ (i)

and $27a-12b=3$ ........... (ii)

On multiplying equation (i) by 3 and equation (ii) by 2 and adding the two equations,

$3(12a+8b)+2(27a-12b)=8\times 3+3\times 2$

$\therefore 36a+24b+54a-24b=24+6$

$\therefore 90a=30\Rightarrow a=\frac{30}{90}=\frac{1}{33}$

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

$12\times \frac{1}{3}+8b=8\therefore 4+8b=8$

$\Rightarrow 8b=4\Rightarrow b=\frac{1}{2}$

$\because \frac{1}{x+y}=a=\frac{1}{3}\Rightarrow x+y=3$ ......(iii)

and $\frac{1}{x+y}=b=\frac{1}{2}\Rightarrow x-y=2$ ......(iv)

On adding equation (iii) and (iv),

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

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

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

$\Rightarrow y=\frac{1}{2}$

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