On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and the y - axis respectively.
Graph of `3x + y - 11 = 0 `
` 3x + y - 11 = 0 rArr y = (11 - 3x ) " " `… (i)
Putting ` x = 2 ` in (i) we get ` y = 5 `
Putting ` x = 3 ` in (i), we get ` y = 2 `
Putting ` x = 5 ` in (i), we get ` y = - 4 `.
Thus, we have the following table for equation (i).
On the graph paper, plot the points `A(2, 5), B(3, 2 ) and C(5, - 4)`.
Joing AB and BC to get the graph line ABC.
Thus, the line ABC is the graph of the equation ` 3x + y - 11 = 0 `
Graph of `x - y - = 0 `
` x - y - 1 = 0 rArr y = (x - 1 )" " `...(ii)
Putting ` x = - 3 ` in (ii), we get ` y = - 4 `
Putting ` x = 0 ` in (ii), we get `y = - 1 `
Putting ` x = 3 `in (ii), we get `y = 2 `.
Thus, we have the following table for equation (ii).
On same graph paper as above, the plot the points `P (-3, - 4) and Q(0, - 1 )`. The third point `B(3, 2 )` is already plotted.
Join `PQ and QB` to get the graph line PQB.
Thus , line PQB is the graph of the equation ` x - y - 1 = 0`
The two graph lines intersect at the point `B(3, 2 )`
` therefore x = 3, y = 2 ` is the solution of the given system of equations.
The region bounded by these lines and y - axis has been shaded.
On extending the graph lines on both sides, we find that these graph lines intersect the y-axis at the points `Q(0, - 1 ) and R(0, 11)`
