Question

solve the following system of linear equations graphically x-y=1,2x+y=8​

Answer 1

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

Given pair of lines are

• x - y = 1 ------(1)

• 2x + y = 8 ------(1).

Consider

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

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

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

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

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

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

$\bf\implies \:x = 1$

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 (0 , - 1) & (1 , 0)

➢ See the attachment graph. Red line

Consider

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

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 (0 , 8) & (4 , 0)

➢ See the attachment graph. Blue line

Hence,

From graph we concluded that

$\purple{\bf :\longmapsto\:Solution \: is \: (3, 2)}$