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

Solution:

GIVEN:

$a x+b y=\frac{a+b}{2}$

$3 x+5 y=4$

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

$a x+b y-\frac{a+b}{2}=0$

$3 x+5 y-4=0$

By cross multiplication method we get

$\frac{x}{(-4 b)-\left(-\frac{5(a+b)}{2}\right)}=\frac{-y}{(-4 a)-\left(-\frac{3(a+b)}{2}\right)}=\frac{1}{5 a-3 b}$

$\frac{x}{(-4 b)+\frac{5(a+b)}{2}}=\frac{-y}{(-4 a)+\frac{3(a+b)}{2}}=\frac{1}{5 a-3 b}$

$\frac{x}{(-4 b)+\frac{5(a+b)}{2}}=\frac{1}{5 a-3 b}$

$x(5 a-3 b)=(-4 b)+\frac{5(a+b)}{2}$

$\Rightarrow x=\frac{1}{2}$

And

$\frac{-y}{(-4 a)+\frac{3(a+b)}{2}}=\frac{1}{5 a-3 b}$

$\frac{-y}{\frac{-8 a+3 a+3 b}{2}}=\frac{1}{5 a-3 b}$

$\frac{-y}{\frac{-5 a+3 b}{2}}=\frac{1}{5 a-3 b}$

$\Rightarrow y(5 a-3 b)=\frac{5 a-3 b}{2}$

$\Rightarrow y=\frac{1}{2}$

Hence we get the value of $x=\frac{1}{2}$ and $\mathrm{y}=\frac{1}{2}$