Solve for x and y:\(\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}4\),\(\frac{

1.	Solve for x and y:\(\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}4\),\(\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}8\)
Answer» The given equations are \(\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}4\).........(i) \(\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}8\) \(\frac{1}{3x+y}-\frac{1}{3x-y}=-\frac{1}4\) (Multiplying by 2) ……(ii) Substituting 1/3x+y = u and 1/3x−y = v in (i) and (ii), we get: u + v = 3/4 ……..(iii) u – v = − 1/4 …….(iv) Adding (iii) and (iv), we get: 2u = 1/2 ⇒ u = 1/4 ⇒3x + y = 4 \((∵\frac1{3x+y}=u)\)........(v) Now, substituting u = 1/4 in (iii), we get: 1/4 + v = 3/4 v = 3/4 - 1/4 ⇒v = 1/2 ⇒ 3x – y = 2 \((∵\frac1{3x-y}=v)\)....(vi) Adding (v) and (vi), we get 6x = 6 ⇒x = 1 Substituting x = 1 in (v), we have 3 + y = 4 ⇒y = 1 Hence, x = 1 and y = 1.

Solve for x and y:\(\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}4\),\(\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}8\)

Answer»

The given equations are

\(\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}4\).........(i)

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

\(\frac{1}{3x+y}-\frac{1}{3x-y}=-\frac{1}4\) (Multiplying by 2) ……(ii)

Substituting 1/3x+y = u and 1/3x−y = v in (i) and (ii), we get:

u + v = 3/4 ……..(iii)

u – v = − 1/4 …….(iv)

Adding (iii) and (iv), we get:

2u = 1/2

⇒ u = 1/4

⇒3x + y = 4 \((∵\frac1{3x+y}=u)\)........(v)

Now, substituting u = 1/4 in (iii), we get:

1/4 + v = 3/4

v = 3/4 - 1/4

⇒v = 1/2

⇒ 3x – y = 2 \((∵\frac1{3x-y}=v)\)....(vi)

Adding (v) and (vi), we get

6x = 6 ⇒x = 1

Substituting x = 1 in (v), we have

3 + y = 4 ⇒y = 1

Hence, x = 1 and y = 1.