Question

Solve the following systems of equations:

${\displaystyle \frac{44}{x+y}+\frac{30}{x-y}=10}$

${\displaystyle \frac{55}{x+y}+\frac{40}{x-y}=13}$

Answer 1

Let ${\displaystyle \frac{1}{x+y}=u\text{and}\frac{1}{x-y}=v}$

Then, the system of the given equations becomes

44u + 30v = 10 ....(i)

55u + 40v = 13 ....(ii)

Multiplying equation (i) by 4 and equation (ii) by 3, we get

176u + 120v = 40 ...(iii)

165u + 120v  = 39 ...(iv)

Subtracting equation (iv) by equation (iii), we get

176 - 165u = 40 - 39

=> 11u = 1

${\displaystyle \Rightarrow u=\frac{1}{11}}$

Putting u = 1/11 in equation (i) we get

${\displaystyle 44\times \frac{1}{11}+30v=10}$

4 + 30v = 10

=> 30v = 10 - 4

=> 30v = 6

${\displaystyle \Rightarrow v=\frac{6}{30}=\frac{1}{5}}$

Now ${\displaystyle u=\frac{1}{x+y}}$

${\displaystyle \Rightarrow \frac{1}{x+y}=\frac{1}{11}}$

=> x + y = 11 ...(v)

Adding equation (v) and (vi), we get

2x = 11 + 5

=> 2x = 16

${\displaystyle \Rightarrow x=\frac{16}{2}=8}$

Putting x = 8 in equation (v) we get

8 + y = 11

=> y = 11 - 8 - 3

Hence, solution of the given system of equations is x = 8, y = 3