Question

Solve for x and y:
`2/(3x+2y) + 3/(3x−2y) = 17/5, 5/(3x+2y) + 1/(3x−2y) = 2`

Answer 1

The given equations are
`2/(3x+2y) + 3/(3x−2y) = 17/5`                                       ……(i)
`5/(3x+2y) + 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)=u)`                                      …..(v)
Now, substituting u = `1/5` in (iv), we get
1 + v = 2 ⇒ v = 1
⇒ 3x – 2y = 1   `(∵1/(3x−2y )=v)`                                      …….(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.