Question

Question 1 (vii)
Solve the following pairs of equations by reducing them to a pair of linear equations:
$\frac{10}{x+y}+\frac{2}{x-y}=4$
$\frac{15}{x+y}-\frac{5}{x-y}=-2$

Answer 1

$\frac{10}{x+y}+\frac{2}{x-y}=4$
$\frac{15}{x+y}-\frac{5}{x-y}=-2$
$\frac{10}{x+y}+\frac{2}{x-y}=4$
$\frac{15}{x+y}-\frac{5}{x-y}=-2$
$Putting\frac{1}{x+y}=pand\frac{1}{x-y}=q\text{in the given equations, we get;}$
$10p+2q=4$
$\Rightarrow 10p+2q-4=\mathrm{0...}\left(i\right)$
$15p-5q=-2$
$\Rightarrow 15p-5q+2=\mathrm{0...}\left(ii\right)$
$\text{Using cross multiplication, we get}$
$\frac{p}{4-20}=\frac{q}{-60-\left(20\right)}=\frac{1}{-50-30}$
$\frac{p}{-16}=\frac{q}{-80}=\frac{1}{-80}$
$\frac{p}{-16}=\frac{1}{-80}and\frac{q}{-80}=\frac{1}{-80}$
$p=\frac{1}{5}andq=1$
$p=\frac{1}{x+y}=\frac{1}{5}andq=\frac{1}{x-y}=1$
$x+y=\mathrm{5...}\left(iii\right)andx-y=\mathrm{1...}\left(iv\right)$
$\text{Adding equation (iii) and (iv), we get}$
$2x=6\Rightarrow x=3$
Putting x = 3 in (iii), we get
$3+y=5\Rightarrow y=2$
$\text{Hence, x = 3 and y = 2}$