Question

Show the following system of equation by graphing. identify the type of system 2x + y = -8, x-y = -1​

Answer 1

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

Given pair of linear equations is

$\rm :\longmapsto\:2x + y = - 8$

and

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

Consider, first line

$\rm :\longmapsto\:2x + y = - 8$

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

$\rm :\longmapsto\:2 \times 0+ y = - 8$

$\rm :\longmapsto\: 0+ y = - 8$

$\bf\implies \:y = - 8$

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

$\rm :\longmapsto\:2x + 0= - 8$

$\rm :\longmapsto\:2x= - 8$

$\bf\implies \:x = - \: 4$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 8 \\ \\ \sf 4 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points

➢ See the attachment graph.

Now, Consider second equation

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

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

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

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

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

$\bf\implies \:y = 1$

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

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

$\bf :\longmapsto\:x= - \: 1$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points.

➢ See the attachment graph.

So, From graph we concluded that, the given pair of system of equations is consistent having unique solution and solution is given by x = - 3 and y = - 2.