Solve the following systems of equations graphically:
$2 x-3 y+13=0$
$3 x-2 y+12=0$
The given equations are:
$2 x-3 y+13=0$ $.(i)$
$3 x-2 y+12=0$ ..(ii)
Putting $x=0$ in equation $(i)$, we get:
$\Rightarrow 2 \times 0-3 y=-13$
$\Rightarrow y=13 / 3$
$x=0, \quad y=13 / 3$
Use the following table to draw the graph.
Draw the graph by plotting the two points $A(0,13 / 2)$ and $B(-13 / 2,0)$ from table.
Graph of the equation.... (ii):
$3 x-2 y=-12$ (ii)
Putting $x=0$ in equation $($ ii $)$ we get:
$\Rightarrow 3 \times 0-2 y=-12$
$\Rightarrow y=6$
$x=0, \quad y=6$
Putting $y=0$ in equation $(i i)$, we get:
$\Rightarrow 3 x-2 \times 0=-12$
$\Rightarrow x=-4$
$x=-4, \quad y=0$
Use the following table to draw the graph.
Draw the graph by plotting the two points $C(0,6)$ and $D(-4,0)$ from table.
The two lines intersect at points $P(-2,3)$.
Hence, $x=-2$ and $y=3$ is the solution.
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