Question

Solve the following pair of equations by reducing them to a pair of linear equations:

$\frac{10}{(x+y)}+\frac{2}{(x-y)}=4,\frac{15}{(x+y)}-\frac{5}{(x-y)}=-2$

• A. $x=1,y=3$
• B. $x=3,y=2$
• C. $x=8,y=1$
• D. $x=0,y=3$

Answer 1

The correct option is B $x=3,y=2$

Given:

$\frac{10}{x+y}+\frac{2}{x-y}=4$

$\frac{15}{x+y}+\frac{5}{x-y}=-2$

let $\frac{1}{x+y}=aand\frac{1}{x-y}=b$

$10a+2b=\mathrm{4.......}eq1$
$15a-5b=-\mathrm{2........}eq2$

Multiply eq1 with 3 and eq2 with 2 we get
$30a+6b=\mathrm{12......}eq3$
$30a-10b=-\mathrm{4......}eq4$

subtracting eq3 and eq4 we get
$16b=16$
$b=1$

Substituting $b=1$ in eq1 we get
$10a+2\times 1=4$
$10a=2$
$a=\frac{1}{5}$

As $\frac{1}{x+y}=a=\frac{1}{5}$

$x+y=\mathrm{5............}eq5$

and $\frac{1}{x-y}=b=1$

$x-y=\mathrm{1...........}eq6$

subtracting eq5 and eq6 we get
$2x=6$
$x=3$

putting $x=3$ in eq 5 we get
$3+y=5$
$y=2$

hence $x=3$ and $y=2$