Question

Solve the following system of linear equations graphically .
a) 2x + 3y = 12, 2y - 1 = x
Please answer I need it urgently.
If you don't know do not give suggestions because I only need answer not your suggestions.

Answer 1

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

Given pair of linear equations is

$\rm :\longmapsto\:2x + 3y = 12$

and

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

Now, Consider

$\rm :\longmapsto\:2x + 3y = 12$

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

$\rm :\longmapsto\:2(0) + 3y = 12$

$\rm :\longmapsto\:3y = 12$

$\bf\implies \:y = 4$

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

$\rm :\longmapsto\:2x + 3(0) = 12$

$\rm :\longmapsto\:2x = 12$

$\bf\implies \:x = 6$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points (0 , 4) & (6 , 0)

➢ See the attachment graph.

Now, Consider

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

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

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

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

$\bf\implies \:y = 0.5$

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

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

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

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

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0.5 \\ \\ \sf - 1 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 0.5) & (- 1 , 0)

➢ See the attachment graph.

From graph we concluded that

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: Solution \: is \: \begin{cases} &\sf{x \: = \: 3} \\ \\ &\sf{y \: = \: 2} \end{cases}\end{gathered}\end{gathered}$

Answer 2

Answer:

Solution−

Given pair of linear equations is

\rm :\longmapsto\:2x + 3y = 12:⟼2x+3y=12

and

\rm :\longmapsto\:2y - 1 = x:⟼2y−1=x

Now, Consider

\rm :\longmapsto\:2x + 3y = 12:⟼2x+3y=12

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

\rm :\longmapsto\:2(0) + 3y = 12:⟼2(0)+3y=12

\rm :\longmapsto\:3y = 12:⟼3y=12

\bf\implies \:y = 4⟹y=4

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

\rm :\longmapsto\:2x + 3(0) = 12:⟼2x+3(0)=12

\rm :\longmapsto\:2x = 12:⟼2x=12

\bf\implies \:x = 6⟹x=6

Hᴇɴᴄᴇ,

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

\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 4 \\ \\ \sf 6 & \sf 0 \end{array}} \\ \end{gathered}\end{gathered}x06y40

➢ Now draw a graph using the points (0 , 4) & (6 , 0)

➢ See the attachment graph.

Now, Consider

\rm :\longmapsto\:2y - 1 = x:⟼2y−1=x

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

\rm :\longmapsto\:2y - 1 = 0:⟼2y−1=0

\rm :\longmapsto\:2y = 1:⟼2y=1

\bf\implies \:y = 0.5⟹y=0.5

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

\rm :\longmapsto\:2(0) - 1 = x:⟼2(0)−1=x

\rm :\longmapsto\:0 - 1 = x:⟼0−1=x

\bf\implies \:x = - 1⟹x=−1

Hᴇɴᴄᴇ,

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

\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0.5 \\ \\ \sf - 1 & \sf 0 \end{array}} \\ \end{gathered}\end{gathered}x0−1y0.50

➢ Now draw a graph using the points (0 , 0.5) & (- 1 , 0)

➢ See the attachment graph.

From graph we concluded that

\begin{gathered}\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: Solution \: is \: \begin{cases} &\sf{x \: = \: 3} \\ \\ &\sf{y \: = \: 2} \end{cases}\end{gathered}\end{gathered}\end{gathered}:⟼Solutionis⎩⎪⎪⎨⎪⎪⎧x=3y=2