3x+y=11 and x-y=1 show it graphically …

On a graph paper, draw a horizontal line XOX' and a vertical line YOY' as the x-axis and they-axis respectively.
{tex}3x + y - 11 = 0{/tex} {tex}\Rightarrow{/tex}{tex}y = (11 -3x){/tex}. ...(i)
Putting {tex}x = 2{/tex} in (i), we get {tex}y = 5{/tex}.
Putting {tex}x = 3{/tex} in (i), we get {tex}y = 2{/tex}.
Putting {tex}x = 5{/tex} in (i), we get {tex}y = -4{/tex}.

On the graph paper, plot the points {tex}A (2, 5), B(3, 2)\ and\ C(5, -4).{/tex}
Join AB and BC to get the graph line ABC.
Thus, the line ABC is the graph of the equation {tex}3x + y - 11 = 0{/tex}.
{tex}x - y - 1 = 0{/tex} {tex}\Rightarrow{/tex}{tex}y = (x - 1){/tex} . ...(ii)
Putting {tex}x = -3{/tex} in (ii), we get {tex}y = -4{/tex}.
Putting {tex}x = 0{/tex} in (ii), we get {tex}y = -1{/tex}.
Putting {tex}x = 3{/tex} in (ii), we get {tex}y = 2{/tex}.

On the same graph paper as above, plot the points {tex}P(-3, -4)\ and\ Q(0, -1){/tex}. The third point {tex}B(3,2){/tex} is already plotted.
Join PQ and QB to get the graph line PQB.
Thus, line PQB is the graph of the equation {tex}x - y - 1 = 0{/tex}.
The two graph lines intersect at the point {tex}B(3,2){/tex}. {tex}x = 3, y = 2{/tex} is the solution of the given system of equations.
The region bounded by these lines and the 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).
Area of triangle = {tex}\frac 12 \times 12 \times 3 = 18{/tex} sq units.