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Solve the following system of equations graphically:
Shade the region bounded by the lines and the y-axis.
$4x\ -\ y\ =\ 4, \ 3x\ +\ 2y\ =\ 14$
Given:
The given equations are:
$4x\ -\ y\ =\ 4, \ 3x\ +\ 2y\ =\ 14$
To do:
We have to solve the given system of linear equations and shade the region bounded by the given lines and the y-axis.
Solution:
To represent the above equations graphically we need at least two solutions for each of the equations.
For equation $4x-y=4$,
$y=4x-4$
If $x=1$ then $y=4(1)-4=4-4=0$
If $x=2$ then $y=4(2)-4=8-4=4$
$x$ | $1$ | $2$ |
$y$ | $0$ | $4$ |
For equation $3x+2y=14$,
$2y=14-3x$
$y=\frac{14-3x}{2}$
If $x=4$ then $y=\frac{14-3(4)}{2}=\frac{14-12}{2}=\frac{2}{2}=1$
If $x=2$ then $y=\frac{14-3(2)}{2}=\frac{14-6}{2}=\frac{8}{2}=4$
$x$ | $4$ | $2$ |
$y$ | $1$ | $4$ |
The equation of y-axis is $x=0$.
The above situation can be plotted graphically as below:
The lines AB and CD represent the equations $4x-y=4$ and $3x+2y=14$ respectively.
The shaded area is the region bounded by the given lines and y-axis.