Question

Solve the following pair 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. $x=2,y=7$
• B. $x=5,y=0$
• C. $x=1,y=1$
• D. $x=2,y=3$

Answer 1

The correct option is C $x=1,y=1$

Given:

$\frac{1}{(3x+y)}+\frac{1}{(3x-y)}=\frac{3}{4},\frac{1}{2(3x+y)}+\frac{1}{2(3x-y)}=\frac{-1}{8}$

$let\frac{1}{3x+y}=aand\frac{1}{3x-y}=b$

Then$a+b=\frac{3}{4}\Rightarrow 4a+4b=\mathrm{3........}eq1$

$\frac{a}{2}-\frac{b}{2}=\frac{-1}{8}$

$\frac{a-b}{2}=\frac{-1}{8}$

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

$4a-4b=-\mathrm{1..............}eq2$

Adding eq1 and eq2 we get

$8a=2$

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

Substituting the value of a in eq1 we get

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

$4b=2$

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

Then $\frac{1}{3x+y}=\frac{1}{4}$

$3x+y=\mathrm{4................}eq3$

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

$3x-y=\mathrm{2...............}eq4$

Adding eq3 and eq4 we get
$6x=6$
$x=1$

Substituting $x=1$ in eq3 we get
$3\times 1+y=4$
$y=1$

Thus $x=1$ and $y=1$