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Solve the following system of linear equations graphically and shade the region between the two lines and x-axis:
$2x\ +\ 3y\ =\ 12$
$x\ -\ y\ =\ 1$
Given:
The given system of equations is:
$2x\ +\ 3y\ =\ 12$
$x\ -\ y\ =\ 1$
To do:
We have to solve the given system of equations and shade the region between the two lines and x-axis.
Solution:
The given pair of equations is:
$2x\ +\ 3y\ -\ 12\ =\ 0$....(i)
$3y=12-2x$
$y=\frac{12-2x}{3}$
$x-y-1=0$.....(ii)
$y=x-1$
To represent the above equations graphically we need at least two solutions for each of the equations.
For equation (i),
If $x=3$ then $y=\frac{12-2(3)}{3}=\frac{12-6}{3}=\frac{6}{3}=2$
If $x=0$ then $y=\frac{12-2(0)}{3}=\frac{12}{3}=4$
$x$ | $3$ | $0$ |
$y=\frac{12-2x}{3}$ | $2$ | $4$ |
For equation (ii),
If $x=3$ then $y=3-1=2$
If $x=1$ then $y=1-1=0$
$x$ | $3$ | $1$ |
| $y=x-1$ | $2$ | $0$ |
The above situation can be plotted graphically as below:
The lines AB and CD represent the equations $2x+3y=12$ and $x-y=1$.
The solution of the given system of equations is the intersection point of the lines AB and CD.
Hence, the solution of the given system of equations is $x=3$ and $y=2$.
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