Question

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

Answer 1

The given equations are
${\displaystyle \frac{44}{x+y}+\frac{30}{x-y}=10}$   ……(i)
${\displaystyle \frac{55}{x+y}-\frac{40}{x-y}=13}$   ……(ii)

Putting ${\displaystyle \frac{1}{x+y}=u\text{and}\frac{1}{x-y}=}$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 = ${\displaystyle \frac{1}{11}}$

⇒ ${\displaystyle \frac{1}{x+y}=\frac{1}{11}\Rightarrow x+y=11}$ …….(vii)
On substituting u = ${\displaystyle \frac{1}{11}}$ in (iii), we get:
4 + 30v = 10
⇒30v = 6
⇒v =${\displaystyle \frac{6}{30}=\frac{1}{5}}$

⇒ ${\displaystyle \frac{1}{x-y}=\frac{1}{5}\Rightarrow x–y=5}$        …….(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.