Solve for x and y:\(\frac{44}{x+y}+\frac{30}{x-y}=10\),\(\frac{55}{x+y

1.	Solve for x and y:\(\frac{44}{x+y}+\frac{30}{x-y}=10\),\(\frac{55}{x+y}-\frac{40}{x-y}=13\)
Answer» The given equations are \(\frac{44}{x+y}+\frac{30}{x-y}=10\)............(i) \(\frac{55}{x+y}-\frac{40}{x-y}=13\)..........(ii) Putting 1/x+y = u and 1/x−y = v, we get: 44u + 30v = 10 ……..(iii) 55u + 40v = 13 …….(iv) On multiplying (iii) by 4 and (iv) by 3, we get: 176u + 120v = 40 …..(v) 165u + 120v = 39 …..(vi) On subtracting (vi) and (v), we get: 11u = 1 ⇒u = 1/11 ⇒ 1/x+y = 1/11 ⇒ x + y = 11 …….(vii) On substituting u = 1/11 in (iii), we get: 4 + 30v = 10 ⇒30v = 6 ⇒v = 6/30 = 1/5 ⇒ 1/x−y = 1/5 ⇒ x – y = 5 …….(viii) On adding (vii) and (viii), we get 2x = 16 ⇒ x = 8 On substituting x =8 in (vii), we get: 8 + y = 11 ⇒ y = 11 – 8 = 3 Hence, the required solution is x = 8 and y = 3.

Solve for x and y:\(\frac{44}{x+y}+\frac{30}{x-y}=10\),\(\frac{55}{x+y}-\frac{40}{x-y}=13\)

Answer»

The given equations are

\(\frac{44}{x+y}+\frac{30}{x-y}=10\)............(i)

\(\frac{55}{x+y}-\frac{40}{x-y}=13\)..........(ii)

Putting 1/x+y = u and 1/x−y = v, we get:

44u + 30v = 10 ……..(iii)

55u + 40v = 13 …….(iv)

On multiplying (iii) by 4 and (iv) by 3, we get:

176u + 120v = 40 …..(v)

165u + 120v = 39 …..(vi)

On subtracting (vi) and (v), we get:

11u = 1

⇒u = 1/11

⇒ 1/x+y = 1/11 ⇒ x + y = 11 …….(vii)

On substituting u = 1/11 in (iii), we get:

4 + 30v = 10

⇒30v = 6

⇒v = 6/30 = 1/5

⇒ 1/x−y = 1/5 ⇒ x – y = 5 …….(viii)

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

2x = 16

⇒ x = 8

On substituting x =8 in (vii), we get:

8 + y = 11

⇒ y = 11 – 8 = 3

Hence, the required solution is x = 8 and y = 3.