Question

Solve the following systems of equations by the method of cross-multiplication:
$\frac{57}{x+y}+\frac{6}{x-y}=5$

$\frac{38}{x+y}+\frac{21}{x-y}=9$

Answer 1

$\frac{57}{(x+y)}+\frac{6}{(x-y)}=5$

and$,\frac{38}{(x+y)}+\frac{21}{(x-y)}=q$

Let$,\frac{1}{x+y}=p$

and$,\frac{1}{x-y}=q\therefore 57p+6q=5⟶\left(i\right)38p+21q=9⟶\left(ii\right)$

On cross-multiplication ,

$\frac{p}{\left|\begin{array}{cc}6& -5\\ 21& -9\end{array}\right|}=\frac{q}{\left|\begin{array}{cc}-5& 57\\ -9& 38\end{array}\right|}=\frac{1}{\left|\begin{array}{cc}57& 6\\ 38& 21\end{array}\right|}\Rightarrow \frac{p}{-54+105}=\frac{q}{-190+153}=\frac{1}{1197-228}\therefore p=\frac{51}{969}=\frac{1}{19}$

and$,q=\frac{322}{969}=\frac{1}{3}\therefore x+y=19⟶\left(iii\right)\{\therefore p=\frac{1}{(x+y)}\}$

and$,x-y=3⟶\left(iv\right)\{\therefore q=\frac{1}{x-y}\}$

Onadding$:2x=22\therefore x=11$

Putting the value of $x$ equation$⟶\left(iii\right)$

$11+y=19$

$\therefore y=8$