Question

Solve the following system of equations graphically:

2x – 3y + 13 = 0 

3x – 2y + 12 = 0

Answer 1

Given, 

2x – 3y + 13 = 0……. (i) 

3x – 2y + 12 = 0……. (ii) 

For equation (i), 

⇒ y = \(\frac{(2x + 13)}{3}\) 

When x = -5, we have y = \(\frac{(2(-5) + 13))}{3}\) = 1 

When x = -2, we have y = \(\frac{(2(-2) + 13))}{3}\) = 3 

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

x	-5	-2
y	1	3

For equation (ii), 

We solve for y: 

⇒ y = \(\frac{(3x + 12)}{2}\) 

So, when x = -4 

y = \(\frac{(3(-4) + 12)}{2}\) = 0 

And, when x = -2 

⇒ y = \(\frac{(3(-2) + 12)}{2}\) = 3 

Thus we have the following table giving points on the line 3x – 2y + 12 = 0

x	-4	-2
y	0	3

Graph of the equations (i) and (ii) is as below:

Clearly the two lines intersect at a single point P (-2, 3) 

Hence, x= -2 and y = 3