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Solve graphically the following system of linear equations. Also, find the coordinates of the points where the lines meet the axis of x:
$2x\ +\ 3y\ =\ 8$
$x\ -\ 2y\ =\ -3$
Given:
The given system of equations is:
$2x\ +\ 3y\ =\ 8$
$x\ -\ 2y\ =\ -3$
To do:
We have to solve the given system of equations and find the coordinates of the points where the lines meet axis of x.
Solution:
The given pair of equations is:
$2x+3y-8=0$....(i)
$3y=8-2x$
$y=\frac{8-2x}{3}$
$x-2y+3=0$.....(ii)
$2y=x+3$
$y=\frac{x+3}{2}$
To represent the above equations graphically we need at least two solutions for each of the equations.
For equation (i),
If $x=4$ then $y=\frac{8-2(4)}{3}=\frac{8-8}{3}=0$
If $x=1$ then $y=\frac{8-2(1)}{3}=\frac{8-2}{3}=\frac{6}{3}=2$
$x$ | $4$ | $1$ |
$y$ | $0$ | $2$ |
For equation (ii),
If $x=-3$ then $y=\frac{-3+3}{2}=\frac{0}{2}=0$
If $x=1$ then $y=\frac{1+3}{2}=\frac{4}{2}=2$
$x$ | $-3$ | $1$ |
$y$ | $0$ | $2$ |
The above situation can be plotted graphically as below:
The lines AB and CD represent the equations $2x+3y=8$ and $x-2y=-3$.
The solution of the given system of equations is the intersection point of the lines AB and CD and these lines meet X-axis at points A and C respectively.
Hence, the solution of the given system of equations is $x=1$ and $y=2$. The lines represented by the equations $2x+3y=8$ and $x-2y=-3$ meet X-axis at $(4,0)$ and $(-3,0)$ respectively.