Question

Draw the graph of following equations: 2x - y = 1, x + 2y = 13 and 1.) find the solution of the equations from the graph. 2.) shade the triangular region formed by the lines and the y - axis.

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Answer 1

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

The given pair of lines is

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

and

$\rm :\longmapsto\:x + 2y = 13$

Now, Consider Line 1

$\red{\rm :\longmapsto\:2x - y = 1}$

can be rewritten as

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

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

$\rm :\longmapsto\:y = 2(0) - 1$

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

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

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

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

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

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

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

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

$\rm :\longmapsto\:y = 4 - 1$

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

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 1 \\ \\ \sf 1 & \sf 1 \\ \\ \sf 2 & \sf 3 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.

Now, Consider Line 2

$\red{\rm :\longmapsto\:x + 2y = 13}$

can be rewritten as

$\rm :\longmapsto\:x = 13 - 2y$

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

$\rm :\longmapsto\:x = 13 - 2(0)$

$\rm :\longmapsto\:x = 13 - 0$

$\rm :\longmapsto\:x = 13$

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

$\rm :\longmapsto\:x = 13 - 2(1)$

$\rm :\longmapsto\:x = 13 - 2$

$\rm :\longmapsto\:x = 11$

Substituting 'y= 4' in the given equation, we get

$\rm :\longmapsto\:x = 13 - 2(4)$

$\rm :\longmapsto\:x = 13 - 8$

$\rm :\longmapsto\:x = 5$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 13 & \sf 0 \\ \\ \sf 11 & \sf 1 \\ \\ \sf 5 & \sf 4 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.

So, from graph we concluded that

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:Solution-\begin{cases} &\sf{x = 3} \\ \\ &\sf{y = 5} \end{cases}\end{gathered}\end{gathered}$

And Required triangle is formed by two lines with y axis is ABC having coordinates

Coordinates of A (0, 6.5)

Coordinates of C (3, 5)

Coordinates of B (0, - 1)