Solve the following system of equations by the method of cross-multiplication:
$x\ +\ 2y\ +\ 1\ =\ 0$
$2x\ –\ 3y\ –\ 12\ =\ 0$

Given: 

The given system of equations is:

$x\ +\ 2y\ +\ 1\ =\ 0$

$2x\ –\ 3y\ –\ 12\ =\ 0$

 To do: 

Here, we have to solve the given system of equations by the method of cross-multiplication.

Solution:  

The solution of a linear pair(standard form) of equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ is given by,

$\frac{x}{b_1c_2-b_2c_1}=\frac{-y}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}$

Comparing the given equations with the standard form of the equations, we get,

$a_1=1, b_1=2, c_1=1$ and $a_2=2, b_2=-3, c_2=-12$

Therefore,

$\frac{x}{2\times(-12)-(-3)\times1}=\frac{-y}{1\times(-12)-2\times1}=\frac{1}{1\times(-3)-2\times2}$

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

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

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

$x=\frac{-21\times1}{-7}$ and $-y=\frac{-14\times1}{-7}$

$x=\frac{-21}{-7}$ and $-y=\frac{-14}{-7}$

$x=3$ and $-y=2$

$x=3$ and $y=-2$

The solution of the given system of equations is $x=3$ and $y=-2$.

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Updated on: 10-Oct-2022

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