Solve the following system of equation by the method of cross-multiplication:

$2(\mathrm{ax}-\mathrm{by})+a+4b=0 ----------\left(1\right)$$2(\mathrm{bx}+\mathrm{ay})+b-4a=0 ----------\left(2\right)$$$$\mathrm{Multiply} \mathrm{equation} \left(1\right) \mathrm{by} a$$\mathrm{and} \mathrm{equation} \left(2\right) \mathrm{by} b$$\mathrm{then} \mathrm{add} \mathrm{the} \mathrm{equations},$$x(2a^2+2b^2)+(a^2+b^2)=0$$\mathrm{hence} x = \frac{-(a^2+b^2)}{2(a^2+b^2)}=\frac{-1}{2}$$\mathrm{putting} \mathrm{in} \mathrm{equation} \left(1\right)$$2\mathrm{by} = 2\mathrm{ax}+a+4b = 4b$$\mathrm{hence} y = 2$">