Question

solve the following equation graphically:3x-2y=4,5x-2y=0​

Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\:3x - 2y = 4$

and

$\rm :\longmapsto\:5x - 2y = 0$

Consider first Line

$\red{\rm :\longmapsto\:3x - 2y = 4}$

$\red{\rm :\longmapsto\:- 2y = 4 - 3x}$

$\red{\rm :\longmapsto\:2y = 3x - 4}$

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

$\red{\rm :\longmapsto\:2y = 3 \times 0 - 4}$

$\red{\rm :\longmapsto\:2y = 0 - 4}$

$\red{\rm :\longmapsto\:2y = - 4}$

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

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

$\red{\rm :\longmapsto\:2y = 3 \times 2 - 4}$

$\red{\rm :\longmapsto\:2y = 6 - 4}$

$\red{\rm :\longmapsto\:2y = 2}$

$\red{\rm :\longmapsto\:y = 1}$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points.

➢ See the attachment graph.

Now, Consider Second Line

$\purple{\rm :\longmapsto\:5x - 2y = 0}$

$\purple{\rm :\longmapsto\:- 2y = - 5x}$

$\purple{\rm :\longmapsto\:2y = 5x}$

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

$\purple{\rm :\longmapsto\:2y = 5 \times 0}$

$\purple{\rm :\longmapsto\:2y = 0}$

$\purple{\rm :\longmapsto\:y = 0}$

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

$\purple{\rm :\longmapsto\:2y = 5 \times 2}$

$\purple{\rm :\longmapsto\:y = 5}$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points.

➢ See the attachment graph.

Hence, From graph we concluded that

$\red{\begin{gathered}\begin{gathered}\bf\: Solution-\begin{cases} &\sf{x \: = \: - \: 2} \\ \\ &\sf{y = \: - \: 5} \end{cases}\end{gathered}\end{gathered}}$