Question

Solve the following systems of equations:

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

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

Answer 1

The given equation is

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

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

Let ${\displaystyle \frac{1}{3x+y}=u\text{and}\frac{1}{3x-y}=v}$ then equation are

${\displaystyle u+v=\frac{3}{4}}$ ...(i)

${\displaystyle \frac{u}{2}-\frac{v}{2}=\frac{1}{8}}$ ....(ii)

Multiply equation (ii) by 2 and add both equations, we get

${\displaystyle u+v=\frac{3}{4}}$

${\displaystyle u-v=-\frac{1}{4}}$

${\displaystyle 2u=\frac{1}{2}}$

${\displaystyle u=\frac{1}{4}}$

Put the value of u  in equation (i) we get

${\displaystyle 1\times \frac{1}{4}+v=\frac{3}{4}}$

${\displaystyle v=\frac{1}{2}}$

Then

${\displaystyle \frac{1}{3x+y}=\frac{1}{4}}$   ....(iii)

3x  + y = 4

${\displaystyle \frac{1}{3x-y}=\frac{1}{2}}$  ...(iv)

3x - y = 2

Add both equation we get

3x + y = 4

3x - y = 2

_________

6x = 6

x = 1

Put the value of x in equation (iii) we get

3 x 1 + y = 4

y = 1

Hence value of x = 1 and  y = 1