Question

Solve the following systems of equations by the method of cross-multiplication:
$6(ax+by)=3a+2b$
$6(bx-ay)=3a-2b$

Answer 1

Given equations:-

$6(ax+by)=3a+2b=>6ax+6by-(3a+2b)=06(bx-ay)=3a-2b=>6bx-6ay-(3a-2b)=0$

Using cross-multiplication method,

$\frac{x}{\left(6b\right)(-(3a-2b))-(-6a)(-(3a+2b))}=\frac{y}{(-(3a+2b))\left(6b\right)-(-(3a-2b))\left(6a\right)}=\frac{1}{\left(6a\right)(-6a)-36b^2}\frac{x}{-6(3a^2-2b^2+5ab)}=\frac{y}{6(3a^2-2b^2-5ab)}=\frac{1}{-36(a^2+b^2)}$

$=>\frac{x}{-6(3a^2-2b^2+5ab)}=\frac{1}{-36(a^2+b^2)}or,x=\frac{3a^2-2b^2+5ab}{6(a^2+b^2)}=>\frac{y}{6(3a^2-2b^2-5ab)}=\frac{1}{-36(a^2+b^2)}or,y=\frac{-(3a^2-2b^2-5ab)}{6(a^2+b^2)}\therefore x=\frac{3a^2-2b^2+5ab}{6(a^2+b^2)}andy=\frac{-(3a^2-2b^2-5ab)}{6(a^2+b^2)}$