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 {tex}4x – 5y – 20 = 0{/tex}{tex}4x – 5y – 20 = 0 {/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5y = (4x – 20){/tex}{tex}\\Rightarrow \\quad y = \\frac { ( 4 x – 20 ) } { 5 }{/tex}…..(i)Table for {tex}4x – 5y – 20 = 0.{/tex}\tx025y-4-2.40\tNow, plot the points {tex}A(0, -4),\\ B(2, -2.4)\\ and\\ C(5, 0){/tex} 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 {tex}4x – 5y – 20 = 0.{/tex}{tex}3x\xa0+ 5y -15 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}5y = (15 -3x){/tex}{tex}\\Rightarrow \\quad y = \\frac { ( 15 – 3 x ) } { 5 }{/tex}………(ii)Table for {tex}3x\xa0+ 5y -15 = 0.{/tex}\tx-505y630\tOn the same graph paper as above, plot the points P (-5, 6) and Q(0, 3).The third point C(5, 0) has already been plotted.Join PQ and QC to get the graph line PQC. Extend it on both ways.Thus, the line PQC is the graph of 3x + 5y -15 = 0.The two graph lines intersect at the point C(5,0).{tex}\\therefore{/tex}\xa0x = 5, y = 0 is the solution of the given system of equations.Clearly, the given equations are represented by the graph lines ABC and PQC respectively.The vertices of {tex}\\triangle{/tex}AQC formed by these lines and the y-axis are A(0, -4), Q(0,3) and C(5,0).