Question

Solve the following systems of equations by the method of cross-multiplication:
$2(ax-by)+a+4b=0$
$2(bx+ay)+b-4a=0$

Answer 1

Given pair of equations are $:$

$2(ax-by)+(a+4b)=0$

And $2(bx+ay)+(b-4a)=0$

Which can be written as $:$

$2ax-2by+(a+4b)=o$

and $2bx+2ay+(b-4a)=0$

By cross multiplication method $,$

$\frac{x}{\left|\begin{array}{cc}{b}_1& c_1\\ b_2& c_2\end{array}\right|}=\frac{y}{\left|\begin{array}{cc}{c}_1& a_1\\ c_2& a_2\end{array}\right|}=\frac{1}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}\frac{x}{\left|\begin{array}{cc}-2b& a+4b\\ 2a& b-4a\end{array}\right|}=\frac{y}{\left|\begin{array}{cc}a+4b& 2a\\ b-4a& 2b\end{array}\right|}=\frac{1}{\left|\begin{array}{cc}2a& -2b\\ 2b& 2a\end{array}\right|}\Rightarrow \frac{x}{-2b(b-4a)-2a(a+4b)}=\frac{y}{2b(a+4b)-2a(b-4a)}=\frac{1}{4a^2+4b^2}\Rightarrow \frac{x}{-2(a^2+b^2)}=\frac{y}{8(a^2+b^2)}=\frac{1}{4(a^2+b^2)}\Rightarrow x=\frac{-2(a^2+b^2)}{4(a^2+b^2)}$

and $y=\frac{8(a^2+b^2)}{4(a^2+b^2)}\therefore x=-\frac{1}{2}$

and $y=2.$