Question

Solve the following systems of equations by the method of cross-multiplication:

$ax+by=\frac{a+b}{2}$

$3x+5y=4$

Answer 1

Given equations:-

$ax+by=\frac{a+b}{2}=>2ax+2by-(a+b)=03x+5y=4=>3x+5y-4=0$

Using cross-multiplication method,

$\frac{x}{\left(2b\right)(-4)-\left(5\right)(-(a+b))}=\frac{y}{(-(a+b))\left(3\right)-(-4)\left(2a\right)}=\frac{1}{\left(2a\right)\left(5\right)-\left(3\right)\left(2b\right)}\frac{x}{-8b+5a+5b}=\frac{y}{-3a-3b+8a}=\frac{1}{10a-6b}\frac{x}{5a-3b}=\frac{y}{5a-3b}=\frac{1}{2(5a-3b)}=>\frac{x}{5a-3b}=\frac{1}{2(5a-3b)}or,x=\frac{1}{2}=>\frac{y}{5a-3b}=\frac{1}{2(5a-3b)}or,y=\frac{1}{2}\therefore x=\frac{1}{2}andy=\frac{1}{2}$