Question

Solve the following pairs of equations by reducing them to a pair of linear equations:
$\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}$
$\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{-1}{8}$

• A. (2, 1)
• B. (1, -1)
• C. (3, 2)
• D. (1, 1)

Answer 1

The correct option is D

(1, 1)

$\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}$
$\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{-1}{8}$
Substituting $\frac{1}{3x+y}=pand\frac{1}{3x-y}=q\text{in the given equations, we get}$
$p+q=\frac{3}{4}...\left(i\right)$
$\frac{p}{2}-\frac{q}{2}=-\frac{1}{8}$
$p-q=-\frac{1}{4}...\left(ii\right)$
$\text{Adding (i) and (ii), we get}$
$2p=\frac{3}{4}-\frac{1}{4}$
$2p=\frac{1}{2}$
$p=\frac{1}{4}$
$\text{Substituting the value in equation (ii), we get}$
$\frac{1}{4}-q=-\frac{1}{4}$
$q=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
$p=\frac{1}{3x+y}=\frac{1}{4}$
$3x+y=\mathrm{4...}\left(iii\right)$
$q=\frac{1}{3x-y}=\frac{1}{2}$
$3x-y=\mathrm{2...}\left(iv\right)$
$\text{Adding equations (iii) and (iv), we get}$
$6x=6$
$x=\mathrm{1...}\left(v\right)$
$\text{Putting the value in equation (iii), we get}$
$3\left(1\right)+y=4$
$y=1$
$\text{Hence, x = 1 and y = 1}$