Solve for x and y:3/x+y + 2/x-y = 23/x+y +2/x-y = 2

1.	Solve for x and y:3/x+y + 2/x-y = 23/x+y +2/x-y = 2
Answer» The given equations are 3/x+y + 2/x-y = 2 ……(i) 9/x+y - 4/x-y = 1 ……(ii) Substituting 1/x+y= u and 1/x-y = v, we get: 3u + 2v = 2 ……..(iii) 9u - 4v = 1 …….(iv) On multiplying (iii) by 2, we get: 6u + 4v = 4 …..(v) On adding (iv) and (v), we get: 15u = 5 ⇒u = 5/15 = 1/3 ⇒ 1/x+y = 1/3 ⇒ x + y = 3 …….(vi) On substituting u = 1/3 in (iii), we get 1 + 2v = 2 ⇒2v = 1 ⇒v = 1/2 ⇒ 1/x-y = 1/2 ⇒ x – y = 2 …….(vii) On adding (vi) and (vii), we get 2x = 5 ⇒ x = 5/2 On substituting x = 5/2 in (vi), we have 5/2 + y = 3 ⇒ y = (3 – 5/2 ) = 1/2 Hence, the required solution is x = 5/2 and y = 1/2 .

Answer»

The given equations are

3/x+y + 2/x-y = 2 ……(i)

9/x+y - 4/x-y = 1 ……(ii)

Substituting 1/x+y= u and 1/x-y = v, we get:

3u + 2v = 2 ……..(iii)

9u - 4v = 1 …….(iv)

On multiplying (iii) by 2, we get:

6u + 4v = 4 …..(v)

On adding (iv) and (v), we get:

15u = 5

⇒u = 5/15 = 1/3

⇒ 1/x+y = 1/3

⇒ x + y = 3 …….(vi)

On substituting u = 1/3 in (iii), we get

1 + 2v = 2

⇒2v = 1

⇒v = 1/2

⇒ 1/x-y = 1/2

⇒ x – y = 2 …….(vii)

On adding (vi) and (vii), we get

2x = 5

⇒ x = 5/2

On substituting x = 5/2 in (vi), we have

5/2 + y = 3

⇒ y = (3 – 5/2 ) = 1/2

Hence, the required solution is x = 5/2 and y = 1/2 .

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