Question

48. Solve the following pair of equations. 10/x+y +2/x-y =4 and 15/x+y - 5/x-y =-2​

Answer 1

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{Values \ of \ x \ and \ y \ are}$

$\bold{\frac{29}{12} \ and \frac{5}{6} \ respectively.}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{The \ given \ equations \ are}$

$\bold{\frac{10}{x+y}+\frac{2}{x-y}=4...(1)}$

$\bold{\frac{15}{x+y}+\frac{5}{x-y}=-2...(2)}$

$\bold{Let \frac{1}{x+y} \ be \ a \ and \frac{1}{x-y} \ be \ b}$

$\bold{,we \ get}$

$\bold{10a+2b=4...(3)}$

$\bold{15a+5b=-2...(4)}$

$\bold{Multiply \ eq(3) \ by \ 3, \ we \ get}$

$\bold{30a+6b=12...(5)}$

$\bold{Multiply \ eq (4) \ by \ 2, \ we \ get}$

$\bold{30a+10b=-4...(6)}$

$\bold{Subtract \ eq(5) \ from \ eq(6), \ we \ get}$

$\bold{4b=-16}$

$\bold{b=-4}$

$\bold{Substituting \ in \ eq(3)}$

$\bold{10a+2(-4)=4}$

$\bold{10a-8=4}$

$\bold{10a=12}$

$\bold{a=\frac{6}{5}}$

$\bold{Resubstituting \ values \ of \ a \ b}$

$\bold{x+y=\frac{5}{6}...(7)}$

$\bold{x-y=4...(8)}$

$\bold{Add \ eq (7) \ and \ eq (8)}$

$\bold{2x=\frac{29}{6}}$

$\bold{x=\frac{29}{12}}$

$\bold{Substituting \ in \ eq (7)}$

$\bold{y=\frac{10}{12}}$

$\bold{y=\frac{5}{6}}$

$\bold\purple{\tt{\therefore{Values \ of \ x \ and \ y \ are}}}$

$\bold\purple{\frac{29}{12} \ and \frac{5}{6} \ respectively.}$