Question

y=2x+2and y=3x-1 show equation in graph​

Answer 1

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

Given pair of line is

$\rm \: y = 2x + 2 - - - (1)$

and

$\rm \: y = 3x - 1 - - - - (2)$

Now, Consider Equation (1)

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

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

$\rm \: y = 2 \times 0 + 2$

$\rm \: y = 0 + 2$

$\rm\implies \:y = 2$

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

$\rm \: y = 2 \times 1 + 2$

$\rm \: y = 2 + 2$

$\rm\implies \:y = 4$

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

$\rm \: y = 2 \times 2 + 2$

$\rm \: y = 4 + 2$

$\rm\implies \:y = 6$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 2 \\ \\ \sf 1 & \sf 4 \\ \\ \sf 2 & \sf 6 \end{array}} \\ \end{gathered}$

Now, Consider Equation (2)

$\rm \: y = 3x - 1$

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

$\rm \: y = 3 \times 0 - 1$

$\rm \: y = 0 - 1$

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

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

$\rm \: y = 3 \times 1 - 1$

$\rm \: y = 3 - 1$

$\rm\implies \:y = 2$

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

$\rm \: y = 3 \times 2 - 1$

$\rm \: y = 6 - 1$

$\rm\implies \:y = 5$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points

➢ See the attachment graph.

From, graph we concluded that, given system of equations is consistent having unique solution and is given by

$\begin{gathered}\begin{gathered}\bf\: \begin{cases} &\sf{x = 3} \\ \\ &\sf{y = 8} \end{cases}\end{gathered}\end{gathered}$