Question

Solve the following system of linear equations graphically: 3x+y-11=0 , x-y-1=0 , shade the region bounded by these lines and y-axis. Find the co-ordinates of the points where the graph lines cut the y-axis.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equations is

$\rm :\longmapsto\:3x + y - 11 = 0$

and

$\rm :\longmapsto\:x - y - 1 = 0$

Consider, Equation 1

$\rm :\longmapsto\:3x + y - 11 = 0$

$\rm :\longmapsto\:y = 11 - 3x$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:y = 11 - 3(0)$

$\rm :\longmapsto\:y = 11 - 0$

$\rm :\longmapsto\:y = 11$

Substituting 'x = 1' in the given equation, we get

$\rm :\longmapsto\:y = 11 - 3(1)$

$\rm :\longmapsto\:y = 11 - 3$

$\rm :\longmapsto\:y = 8$

Substituting 'x = 2' in the given equation, we get

$\rm :\longmapsto\:y = 11 - 3(2)$

$\rm :\longmapsto\:y = 11 - 6$

$\rm :\longmapsto\:y = 5$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 11 \\ \\ \sf 1 & \sf 8 \\ \\ \sf 2 & \sf 5 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.

Now, Consider Equation (2)

$\rm :\longmapsto\:x - y - 1 = 0$

$\rm :\longmapsto\:y = x - 1$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:y = 0 - 1$

$\rm :\longmapsto\:y = - 1$

Substituting 'x = 1' in the given equation, we get

$\rm :\longmapsto\:y = 1 - 1$

$\rm :\longmapsto\:y = 0$

Substituting 'x = 2' in the given equation, we get

$\rm :\longmapsto\:y = 2 - 1$

$\rm :\longmapsto\:y = 1$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 1 \\ \\ \sf 1 & \sf 0 \\ \\ \sf 2 & \sf 1 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.

So, Required triangle bounded by these lines with y - axis is triangle ABC.

From graph we concluded that the line cut the y - axis at

( 0, - 1 ) and ( 0, 11 )

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:Solution \: is \: - \: \begin{cases} &\sf{x \: = \: 3} \\ \\ &\sf{y \: = \: 2} \end{cases}\end{gathered}\end{gathered}}$