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