Solve the following systems of equations graphically:
$3 x+y+1=0$
$2 x-3 y+8=0$
The given equations are
$3 x+y+1=0 \quad \ldots \ldots .(i)$
$2 x-3 y+8=0 \quad \ldots \ldots \ldots($ ii $)$
Putting $x=0$ in equation $(i)$, we get:
$\Rightarrow 3 \times 0+y=-1$
$\Rightarrow y=-1$
$x=0, \quad y=-1$
Putting $y=0$ in equation $(i,$, we get:
$\Rightarrow 3 x+0=-1$
$\Rightarrow x=-1 / 3$
$x=-1 / 3, \quad y=0$
Use the following table to draw the graph.
Draw the graph by plotting the two points $A(0,-1)$ and $B(-1 / 3,0)$ from table.
Graph of the equation. (ii):
$2 x-3 y=-8 \quad$......(ii)
Putting $x=0$ in equation (ii) we ge
$\Rightarrow 2 \times 0-3 y=-8$
$\Rightarrow y=8 / 3$
$x=0, \quad y=8 / 3$
Putting $y=0$ in equation $(i i)$, we get
$\Rightarrow 2 x-3 \times 0=-8$
$\Rightarrow x=-4$
$x=-4, \quad y=0$
Use the following table to draw the graph.
Draw the graph by plotting the two points $C(0,8 / 3)$ and $D(-4,0)$ from table.
The two lines intersect at points $\mathrm{P}(-1,2)$.
Hence $x=-1, \quad y=2$ is the solution.