Question

Solve the following simultaneous equation
$\frac{30}{x-y}+\frac{44}{x+y}=10;\frac{40}{x-y}+\frac{55}{x+y}=13$

Answer 1

Given equations are,

$\frac{30}{x-y}+\frac{44}{x+y}=10$ ...(i)

$\frac{40}{x-y}+\frac{55}{x+y}=13$ ...(ii)

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

$\frac{120}{x-y}+\frac{176}{x+y}=40$ ...(iii)

$\frac{120}{x-y}+\frac{165}{x+y}=39$ ...(iv)

Subtracting eq.(iv) from eq.(iii), we get

$0+\frac{11}{x+y}=1$

$\Rightarrow x+y=11$ ...(v)

Substituting eq.(v) in eq.(i) , we get

$\frac{30}{x-y}+\frac{44}{11}=10$

$\Rightarrow \frac{30}{x-y}=10-4$

$\Rightarrow \frac{30}{x-y}=6$

$\Rightarrow x-y=\frac{30}{6}$

$\Rightarrow x-y=5$ ...(vi)

Adding eq.(v) and eq.(vi), we get

$2x=16$

$\therefore x=16/2=8$

Substituting the value of $x$ in eq.(v)

$8+y=11$

$\Rightarrow y=11-8$

$\therefore y=3$

Hence, $x=8$ and $y=3$