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प्रश्न
Solve for x and y:
`44/(x+y) + 30/(x−y) = 10, 55/(x+y) - 40/(x−y) = 13`
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उत्तर
The given equations are
`44/(x+y) + 30/(x−y) = 10 ` ……(i)
`55/(x+y ) - 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.
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