Question

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

$\frac{2}{x}+\frac{3}{y}=13$

$\frac{5}{x}-\frac{4}{y}=-2$

where $x\ne 0$ and $y\ne 0$

Answer 1

Given equations:-

$\frac{2}{x}+\frac{3}{y}=13and\frac{5}{x}-\frac{4}{y}=-2$

Let $\frac{1}{x}=pand\frac{1}{y}=q$

Thus, we have,

$2p+3q=13=>2p+3q-13=05p-4q=-2=>5p-4q+2=0$

Using cross-multiplication method to solve the equations we have,

$\frac{p}{\left(3\right)\left(2\right)-(-4)(-13)}=\frac{q}{(-13)\left(5\right)-\left(2\right)\left(2\right)}=\frac{1}{\left(2\right)(-4)-\left(5\right)\left(3\right)}\frac{p}{6-52}=\frac{q}{-65-4}=\frac{1}{-8-15}\frac{p}{-46}=\frac{q}{-69}=\frac{1}{-23}=>\frac{p}{-46}=\frac{1}{-23}or,p=2=>\frac{q}{-69}=\frac{1}{-23}or,q=3Now,p=\frac{1}{x}=2=>x=\frac{1}{2}andq=\frac{1}{y}=3=>y=\frac{1}{3}\therefore x=\frac{1}{2}andy=\frac{1}{3}$