Question

Solve the system of equations graphically: 

2x + 3y + 5 = 0, 

3x – 2y – 12 = 0

Answer 1

From the first equation, write y in terms of x

y = − \((\frac{5+2x}3)\)........(i)

Substitute different values of x in (i) to get different values of y

For x = -1, y = - \(\frac{5-2}3\) = - 1

For x = 2, y = - \(\frac{5+4}3\) = - 3

For x = 5, y = - \(\frac{8+10}3\) = - 5

Thus, the table for the first equation (2x + 3y + 5 = 0) is

x	-1	2	5
y	-1	-3	-5

Now, plot the points A (-1, -1), B (2, -3) and C (5, -5) on a graph paper and join them to get the graph of 2x + 3y + 5 = 0. 

From the second equation, write y in terms of x

y = \(\frac{3x-12}2\)......(ii)

Now, substitute different values of x in (ii) to get different values of y

For x = 0, y = \(\frac{0-12}2\) = - 6

For x = 2, y = \(\frac{6-12}2\) = - 3

For x = 4, y = \(\frac{12-12}2\) = 0

So, 

the table for the second equation (3x – 2y – 12 = 0) is

x	0	2	4
y	-6	-3	0

Now, plot the points D (0, -6), E (2, -3) and F (4, 0) on the same graph paper and join D, E and F to get the graph of 3x – 2y – 12 = 0.

From the graph it is clear that, the given lines intersect at (2, -3). 

Hence, the solution of the given system of equation is (2, -3).