Question

Solve the system of equations graphically:

2x – 3y + 13 = 0,

3x – 2y + 12 = 0

Solve graphically:

2x – 3y + 13 = 0; 3x – 2y + 12 = 0

Answer 1

From the first equation, write y in terms of x

y = `(2x + 13)/3`   ...(i)

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

For x = –5, y = `(-10 + 13 )/3 = 1`

For x = 1, y = `(2 + 13)/3 = 5`

For x = 4, y = `(8 + 13)/3 = 7`

Thus, the table for the first equation (2x - 3y + 13 = 0) is

x	–5	1	4
y	1	5	7

Now, plot the points A (–5, 1), B (1, 5) and C (4, 7) on a graph paper and join A, B and C to get the graph of 2x – 3y + 13 = 0.

From the second equation, write y in terms of x

y = `(3x - 12)/2`   ...(ii)

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

For x = –4, y = `(-12 + 12)/2 = 0`

For x = –2, y = `(-6 + 12)/2 = 3`

For x = 0, y = `(0 + 12)/2 = 6`

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

x	–4	–2	0
y	0	3	6

Now, plot the points D (–4, 0), E (–2, 3) and F (0, 6) 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).

Answer 2

Given equations:

2x – 3y + 13 = 0

3x – 2y + 12 = 0

2x – 3y = –13

⇒ `y = (2x + 13)/3`

x	0	– 6.5	1
y	4.3	0	5

And 3x – 2y = –12

⇒ `y = (3x + 12)/2`

x	0	– 4	– 2
y	6	0	3

These lines intersect each other at point (–2, 3)

Hence, x = –2 and y = 3.