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Solve the following system of equations graphically:
$2x\ +\ 3y\ +\ 5\ =\ 0$
$3x\ -\ 2y\ β\ 12\ =\ 0$
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 $y=\frac{-2(2)-5}{3}=\frac{-4-5}{3}=\frac{-9}{3}=-3$
$x$ | $-1$ | $2$ |
$y=\frac{-2x-5}{3}$ | $-1$ | $-3$ |
For equation (ii),
If $x=4$ then $y=\frac{3(4)-12}{2}=\frac{12-12}{2}=0$
If $x=6$ then $y=\frac{3(6)-12}{2}=\frac{18-12}{2}=\frac{6}{2}=3$
$x$ | $4$ | $6$ |
$y=\frac{3x-12}{2}$ | $0$ | $3$ |
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 $y=-3$.