Draw the graph of the equation …

The given equations are
{tex}x = 3{/tex} …(i)
{tex}x = 5{/tex} …(ii)
{tex}2x - y - 4 = 0{/tex}
⇒ {tex}y = 2x - 4{/tex}
{tex}\left. \begin{gathered} \begin{array}{*{20}{c}} {x = 0,then}&{y = 2(0) - 4 = 0 - 4 = - 4} \end{array} \hfill \\ \begin{array}{*{20}{c}} {x = 1,then}&{y = 2(1) - 4 = 2 - 4 = - 2} \end{array} \hfill \\ \begin{array}{*{20}{c}} {x = 2,then}&{y = 2(2) - 4 = 4 - 4 = 0} \end{array} \hfill \\ \begin{array}{*{20}{c}} {x = 3,then}&{y = 2(3) - 4 = 6 - 4 = 2} \end{array} \hfill \\ \begin{array}{*{20}{c}} {\mathop {x = 4,then}\limits_{x = 3} }&{\mathop {y = 2(4) - 4 = 8 - 4 = 4}\limits_{x = 5} } \end{array} \hfill \\ \end{gathered} \right\} \Rightarrow{/tex}

The coordinates of the vertices of the required quadrilateral {tex}PMNB{/tex} are {tex}P(3, 0),\ M(5, 0),\ N(5, 6)\ and\ B(3, 2){/tex}
The quadrilateral formed by these given three lines and x-axis is {tex}PMNB{/tex}. It is trapezium. So, area of the required trapezium
{tex}= \frac { 1 } { 2 } ( B P + M N ) \times P M{/tex}
{tex}= \frac { 1 } { 2 } [ ( 2 - 0 ) + ( 6 - 0 ) ] ( 5 - 3 ){/tex}
{tex}= \frac { 1 } { 2 } \times 8 \times 2 = 8{/tex}square units
{tex}\therefore{/tex} the area of required {tex}PMNB{/tex} = {tex}8\ square\ units.{/tex}