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Solve the following system of equations graphically:
$2x\ β\ 3y\ +\ 13\ =\ 0$
$3x\ β\ 2y\ +\ 12\ =\ 0$
Given:
The given system of equations is:
$2x\ –\ 3y\ +\ 13\ =\ 0$
$3x\ –\ 2y\ +\ 12\ =\ 0$
To do:
We have to represent the above system of equations graphically.
Solution:
The given pair of equations are:
$2x\ -\ 3y\ +\ 13\ =\ 0$....(i)
$3y=2x+13$
$y=\frac{2x+13}{3}$
$3x\ -\ 2y\ +\ 12\ =\ 0$....(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 $x=-2$ then $y=\frac{2(-2)+13}{3}=\frac{-4+13}{3}=\frac{9}{3}=3$
If $x=1$ then $y=\frac{2(1)+13}{3}=\frac{2+13}{3}=\frac{15}{3}=5$
| $x$ | $-2$ | $1$ |
$y=\frac{2x+13}{3}$ | $3$ | $5$ |
For equation (ii),
If $x=0$ then $y=\frac{3(0)+12}{2}=\frac{12}{2}=6$
If $x=-2$ then $y=\frac{3(-2)+12}{2}=\frac{-6+12}{2}=\frac{6}{2}=3$
| $x$ | $-2$ | $0$ |
| $y=\frac{3x+12}{2}$ | $3$ | $6$ |
The above situation can be plotted graphically as below:
The line AB represents the equation $2x-3y+13=0$ 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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