Solve the system of equations graphically: 3x+y+1=0 2x-3y+8=0

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Solve the system of equations graphically:
3x+y+1=0
2x-3y+8=0

Solve the system of equations graphically:
3x+y+1=0
2x-3y+8=0

Solution:

Draw a horizontal line on a graph paper X’OX and a vertical line YOY’ as the $\mathrm{x}$ -axis and y-axis, respectively.
Graph of $3 \mathbf{x}+\mathbf{y}+\mathbf{1}=\mathbf{0}$
$\begin{array}{l} 3 x+y+1=0 \\ \Rightarrow y=(-3 x-1) \dots \dots(i) \end{array}$
Putting $x=0$ , we obtain $y=-1$
Putting $x=-1$ , we obtain $y=2$
Putting $x=1$ , we obtain $y=-4$ .

Therefore, we have the following table for the eq. $3 x+y+1=0$ .

$\begin{tabular}{|r|r|r|c|} \hline $\mathrm{x}$ & 0 & $-1$ & 1 \\ \hline $\mathrm{y}$ & $-1$ & 2 & $-4$ \\ \hline \end{tabular}$

So now, plot the points $\mathrm{A}(0,-1), \mathrm{B}(-1,2)$ and $\mathrm{C}(1,-4)$ on the graph paper. Join $\mathrm{AB}$ and $\mathrm{AC}$ to get the graph line BC. Extend it on both ways.

Therefore, BC is the graph of $3 \mathrm{x}+\mathrm{y}+1=0$
$\text { Graph of } 2 x-3 y+8=0$
$\begin{array}{l} 2 x-3 y+8=0 \\ \Rightarrow 3 y=(2 x+8) \\ \therefore y=\frac{2 x+8}{3}\dots \dots(ii) \end{array}$
Putting $x=-1$ , we obtain $y=2$
Putting $x=2$ , we obtain $y=4$
Putting $x=-4$ , we obtain $y=0$ .

Thus, we have the following table for the eq. $2 \mathrm{x}-3 \mathrm{y}+8=0$ .

$\begin{tabular}{|r|r|r|r|} \hline $\mathrm{x}$ & $-1$ & 2 & $-4$ \\ \hline $\mathrm{y}$ & 2 & 4 & 0 \\ \hline \end{tabular}$

So now, plot the points $\mathrm{P}(2,4)$ and $\mathrm{Q}(-4,0)$ . The point $\mathrm{B}(-1,2)$ has already been plotted. Join $\mathrm{PB}$ and $\mathrm{BQ}$ and extend it on both ways.
Therefore, PQ is the graph of $2 \mathrm{x}+\mathrm{y}-8=0$ .

The two graph lines intersect at B $(-1.2)$ .
$\therefore x=-1$ and $y=2$