Draw the graph of the equations …

Graph of {tex}4x - 5y + 16 = 0{/tex}
{tex}4x - 5y + 16 = 0{/tex} {tex}\Rightarrow{/tex} {tex}5y = 4x + 16{/tex}
{tex}\Rightarrow \quad y = \frac { ( 4 x + 16 ) } { 5 }{/tex}.......(i)
Table for {tex}4x - 5y +16 = 0.{/tex}

Now, plot the points A (-4, 0), B(1, 4) and C(6, 8) 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 4x - 5y +16 = 0.

Graph of {tex}2x + y - 6 =0{/tex}
{tex}2x + y - 6 = 0{/tex} {tex}\Rightarrow{/tex} {tex}y = (6 - 2x){/tex}. ... (ii)
Table for {tex}2x + y - 6 = 0.{/tex}

On the same graph paper as above, plot the points P(0,6), Q(2,2) and R(3,0).
Join PQ and QR to get the graph line PQR. Extend it on both ways.
Thus, the line PQR is the graph of 2x + y - 6 = 0.

The two graph lines ABC and PQR intersect at the point B(1,4). x = 1 and y = 4 is the solution of the given system of equations.
These lines form {tex}\triangle{/tex}BAR with the x-axis, whose vertices are B(l, 4), A (-4, 0) and R(3, 0).
The two graph lines ABC and PQR intersect at the point B(1, 4).
{tex}\therefore{/tex}  x = 1 and y = 4 is the solution of the given system of equations.
These lines form {tex}\triangle{/tex}BAR with the x-axis, whose vertices are B(1, 4), A (-4, 0) and R(3, 0).