On a graph paper, draw a horizontal line X'OX and vertical line YOY' as the x- axis and the y-axis respectively.
Graph of ` 3 x - 2y - 11=0`
`3 x + 2y - 11 = 0 rArr 2y = (11 - 3x )`
` " "rArr y = (( 11 - 3x ))/( 2) " " ` ... (i)
Putting ` x = -1 ` in (i), we get `y =7 `
Putting ` x = 1 ` in (i), we get `y = 4`
Putting ` x = 3` in (i), we get `y = 1`
Thus, we have the following table for ` 3x + 2y -11 = 0`
Now, plot the points ` A(-1, 7), B(1, 4)and C(3, 1)` on the graph paper.
Join AB and BC to get the graph line ABC. Extend it on both ways.
Thus, the line ABC is the graph of ` 3x + 2y-11 = 0`
Graph of ` 2x - 3y + 10 = 0`
` 2x - 3y + 10 = 0 rArr 3y = 2x + 10 rArr y= (( 2x + 10) )/(3)" "`... (ii)
Putting ` x = -2 ` in (ii), we get ` y = 2`.
Putting ` x = 1 `in (ii), we get ` y = 4 `
Putting ` x = 4 `in (ii), we get `y = 6 `
Thus, we have the following table for ` 2x - 3y + 10 = 0`
On the same graph paper as above, plot the points `P(-2, 2) and Q(4, 6)`. The third point `B(1, 4)` has already been plotted.
Now, join PB and BQ to get the graph line PBQ. Extend it on both ways.
Thus, the line ` PBQ ` is the graph of ` 2x - 3y + 10=0 `.
The two graph lines intersect at the point `B(1, 4)`
` therefore x = 1, y = 4` is the solution of the given system of equations.
These graph lines intersect the x-axis has been shaded.
The shaded region is the `DeltaBRS` with `B(1, 4), R(-5, 0) and S((11)/(3), 0)`.
