Solve for x and y:\(\frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5}\),\(\

1.	Solve for x and y:\(\frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5}\),\(\frac{5}{3x+2y}+\frac{1}{3x-2y}=2\)
Answer» The given equations are \(\frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5}\)...........(i) \(\frac{5}{3x+2y}+\frac{1}{3x-2y}=2\).........(ii) Substituting 1/3x+2y = u and 1/3x−2y = v, in (i) and (ii), we get: 2u + 3v = 17/5 ……..(iii) 5u + v = 2 …….(iv) Multiplying (iv) by 3 and subtracting from (iii), we get: 2u – 15u = 17/5 – 6 ⇒ –13u = −13/5 ⇒u = 1/5 ⇒3x + 2y = 5 (∵ 1/3x+2y = ) …..(v) Now, substituting u = 1/5 in (iv), we get 1 + v = 2 ⇒ v = 1 ⇒ 3x – 2y = 1 (∵ 1/3x−2y = ) …….(vi) Adding(v) and (vi), we get: ⇒ 6x = 6 ⇒ x = 1 Substituting x = 1 in (v), we get: 3 + 2y = 5 ⇒ y = 1 Hence, x = 1 and y =1.

Solve for x and y:\(\frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5}\),\(\frac{5}{3x+2y}+\frac{1}{3x-2y}=2\)

Answer»

The given equations are

\(\frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5}\)...........(i)

\(\frac{5}{3x+2y}+\frac{1}{3x-2y}=2\).........(ii)

Substituting 1/3x+2y = u and 1/3x−2y = v, in (i) and (ii), we get:

2u + 3v = 17/5 ……..(iii)

5u + v = 2 …….(iv)

Multiplying (iv) by 3 and subtracting from (iii), we get:

2u – 15u = 17/5 – 6

⇒ –13u = −13/5 ⇒u = 1/5

⇒3x + 2y = 5 (∵ 1/3x+2y = ) …..(v)

Now, substituting u = 1/5 in (iv), we get

1 + v = 2 ⇒ v = 1

⇒ 3x – 2y = 1 (∵ 1/3x−2y = ) …….(vi)

Adding(v) and (vi), we get:

⇒ 6x = 6 ⇒ x = 1

Substituting x = 1 in (v), we get:

3 + 2y = 5 ⇒ y = 1

Hence, x = 1 and y =1.