Question

Solve for x and y:
`1/(2(x+2y)) + 5/(3(3x−2y)) = - 3/2, 1/(4(x+2y)) - 3/(5(3x−2y)) = 61/60` where x + 2y ≠ 0 and 3x – 2y ≠ 0.

Answer 1

The given equations are
`1/(2(x+2y)) + 5/(3(3x−2y)) = - 3/2` ……(i)
`1/(4(x+2y)) - 3/(5(3x−2y)) = 61/60 `……(ii)
Putting `1/(x+2y) = u and 1/(3x−2y)` = v, we get:
`1/2 u + 5/3 v = - 3/2`         ……..(iii)
`5/4 u – 3/5 v = 61/60`         …….(iv)
On multiplying (iii) by 6 and (iv) by 20, we get:
3u + 10v = -9                …..(v)
25u – 12v = `61/3`         …..(vi)
On multiplying (v) by 6 and (vi) by 5, we get
18u + 60v = -54               …….(vii)
125u – 60v = `305/3`               ……..(viii)
On adding(vii) and (viii), we get:
`143u = 305/3 – 54 = (305−162)/3 = 143/3`
`⇒u = 1/3 = 1/(x+2y)`

⇒x + 2y = 3                     …….(ix)
On substituting u = `1/3` in (v), we get:
1 + 10v = -9
⇒10v = -10
⇒v = -1
`⇒1/(3x−2y) = -1 ⇒3x – 2y = -1`                 …….(x)
On adding (ix) and (x), we get:
4x = 2
`⇒x = 1/2`
On substituting x = `1/2` in (x), we get:
`3/2 – 2y = -1`
`2y = (3/2+1) = 5/2`
`y = 5/4`
Hence, the required solution is x = `1/2 and y = 5/4`