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

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

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

Solution:

Draw a horizontal line on a graph paper X’OX and a vertical line YOY’ as the $\mathrm{x}$ -axis and $\mathrm{y}$ -axis, respectively.
$\begin{array}{l} x+2 y+2=0 \\ \Rightarrow 2 y=(-2-x) \\ \therefore y=\frac{-2-x}{2} \dots \dots(i) \end{array}$

Putting $x=-2$ , we obtain $y=0$
Putting $x=0$ , we obtain $y=-1$
Putting $x=2$ , we obtain $y=-2$

So, we have the following table for the equation $\mathrm{x}+2 \mathrm{y}+2=0$ .

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

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

So, $\mathrm{AC}$ is the graph of $\mathrm{x}+2 \mathrm{y}+2=0$
$3 x+2 y-2=0$
$\Rightarrow 2 y=(2-3 x)$
$\therefore y=\frac{2-3 x}{2} \quad \ldots \ldots (ii)$

Putting $x=0$ , we obtain $y=1$
Putting $x=2$ , we obtain $y=-2$
Putting $x=4$ , we obtain $y=-5$

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

$\begin{tabular}{|r|r|r|r|} \hline $\mathrm{x}$ & 0 & 2 & 4 \\ \hline $\mathrm{y}$ & 1 & $-2$ & $-5$ \\ \hline \end{tabular}$

Now, plot the points $\mathrm{P}(0,1)$ and $\mathrm{Q}(4,-5)$ . The point $\mathrm{C}(2,-2)$ has already been plotted. Join $\mathrm{PC}$ and $\mathrm{QC}$ and extend it on both ways.
Thus, $\mathrm{PQ}$ is the graph of $3 \mathrm{x}+2 \mathrm{y}-2=0$ .

The two graph lines intersect at $\mathrm{A}(2,-2)$ .
$\therefore x=2 \text { and } y=-2$