Question

Solve each of the following systems of equations graphically:

$2x-3y+13=0,3x-2y+12=0.$

Answer 1

So we have 2 x-3y + 13=0 and 3 x-2y + 12=0

Now, 2x-3y+13=0

=> x = $\frac{(3y-13)}{2}$

When y=1 then, x=-5

When y=3 then, x=-2

Thus, we have the following table giving points on the line 2x-3y+13=0

X	-5	-2
Y	1	3

Now, 3x-2y+12=0

=> x = $\frac{(2y-12)}{3}$

When y=0, then x=-14

When y=3, then x= -2

Thus, we have the following table giving points on the line x-y+3=0

X	-4	-2
Y	0	3

Graph of the equation 2x-3y+14=0 and 3x-2y+12=0

Clearly, two lines intersect at (-2, 3)

Hence x=-2 and y=3 is the solution of the given system of equations.