Question

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

x + 2y + 1 = 0
2x − 3y − 12 = 0

Answer 1

The given system of equation is

x + 2y + 1 = 0

2x − 3y − 12 = 0

Here

${\displaystyle a_1=1,b_1=2,c_1=1}$

${\displaystyle a_2=2,b_2=-3,c_3=-12}$

By cross-multiplication, we get

${\displaystyle \Rightarrow \frac{x}{2\times (-12)-1\times (-3)}=\frac{-y}{1\times (-12)-1\times 2}=\frac{1}{1\times (-3)-2\times 2}}$

${\displaystyle \Rightarrow \frac{x}{-24+3}=\frac{-y}{-12-2}=\frac{1}{-3-4}}$

${\displaystyle \Rightarrow \frac{x}{-21}=\frac{-y}{-14}=\frac{1}{-7}}$

Now

${\displaystyle \frac{x}{-21}=\frac{1}{-7}}$

${\displaystyle \Rightarrow x=\frac{-21}{7}=3}$

and 

${\displaystyle \frac{-y}{-14}=\frac{1}{-7}}$

${\displaystyle \Rightarrow \frac{y}{14}=\frac{-1}{7}}$

${\displaystyle \Rightarrow y=\frac{-14}{7}=-2}$

Hence, the solution of the given system of equations is x = 3 and y = -2