On a graph paper, draw a horizontal line `X'OX` and a vetical line `YOY' ` as the x-axis and the y-axis respectively.
Graph of ` x = 2y = 3 `
` x + 2y = 3 rArr 2y = (3 - x )`
` " " rArr y = (( 3 - x ))/( 2)" " `... (i)
Putting ` x = - 3 ` in (i), we get `y = 3 `
Putting ` x = - 1 ` in (i), we get ` y = 2 `.
Putting ` x = 1 ` in (i), we get ` y= 1 `
Thus, we have the following table for the equation ` x + 2y = 3`.
Now, plot the points ` A (-3, 3), B ( - 1, 2) and C (1, 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 ` x + 2y = 3 `.
Graph of ` 4x + 3y i= 2 `
` 4x + 3y = 2 rArr 3y = ( 2 - 4x)`
` " " rArr y = (( 2- 4x ))/( 3)" " `... (ii)
Putting ` x= -4 `in (ii), we get ` y = 6.`
Putting ` x = - 1 ` in (ii), we get `y = 2 `.
Putting ` x = 2 ` in (ii) , we get `y = - 2 `
Thus, we have the following table for the equation ` 4x + 3y = 2 `.
Now on the same graph paper as above , plot the poins ` P ( - 4, 6)` and `Q(2, - 2 )`. The point ` B (-1, 2)` has already been plotted. Join PB and BQ to get the line PBQ. Extend it on both ways.
Thus, the line ` PBQ ` is the graph of ` 4x + 3y = 2`.
The two graphs lines intersect at the point `B (1, - 2)`
` therefore x = - 1 and y = 2 ` is the solution of the given system of equations.
