Solve the following system of equations graphically:

$2x\ +\ 3y\ +\ 5\ =\ 0$
$3x\ -\ 2y\ –\ 12\ =\ 0$

Given:

The given system of equations is:

$2x\ +\ 3y\ +\ 5\ =\ 0$

$3x\ -\ 2y\ –\ 12\ =\ 0$

To do:

We have to represent the above system of equations graphically.

Solution:

The given pair of equations are:

$2x\ +\ 3y\ +\ 5\ =\ 0$....(i)

$3y=-2x-5$

$y=\frac{-2x-5}{3}$

$3x\ -\ 2y\ -\ 12\ =\ 0$....(ii)

$2y=3x-12$

$y=\frac{3x-12}{2}$

To represent the above equations graphically we need at least two solutions for each of the equations.

For equation (i),

If $x=-1$ then $y=\frac{-2(-1)-5}{3}=\frac{2-5}{3}=\frac{-3}{3}=-1$

If $x=2$ then $y=\frac{-2(2)-5}{3}=\frac{-4-5}{3}=\frac{-9}{3}=-3$

For equation (ii),

If $x=4$ then $y=\frac{3(4)-12}{2}=\frac{12-12}{2}=0$

If $x=6$ then $y=\frac{3(6)-12}{2}=\frac{18-12}{2}=\frac{6}{2}=3$

The above situation can be plotted graphically as below:

The line AB represents the equation $2x+3y+5=0$ and the line PQ represents the equation $3x-2y-12=0$.

The solution of the given system of equations is the intersection point of the lines AB and PQ.

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

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Updated on: 10-Oct-2022

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