Question

Solve the system of equations graphically:

3x + y + 1 = 0, 2x – 3y + 8 = 0

Solve the following system of equations graphically:

3x + y + 1 = 0, 2x – 3y + 8 = 0

Answer 1

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.

Graph of 3x + y + 1 = 0

3x + y + 1 = 0

⇒ y = (–3x – 1)   ...(i)

Putting x = 0, we get y = –1.

Putting x = –1, we get y = 2.

Putting x = 1, we get y = –4.

Thus, we have the following table for the equation 3x + y + 1 = 0.

x	0	–1	1
y	–1	2	–4

Now, plot the points A(0, –1), B(–1, 2) and C(1, –4) on the graph paper.

Join AB and AC to get the graph line BC. Extend it on both ways.

Thus, BC is the graph of 3x + y + 1 = 0.

Graph of 2x – 3y + 8 = 0

2x – 3y + 8 = 0

⇒ 3y = (2x + 8)

∴ `y = (2x + 8)/3`

Putting x = –1, we get y = 2.

Putting x = 2, we get y = 4.

Putting x = –4, we get y = 0.

Thus, we have the following table for the equation 2x – 3y + 8 = 0.

x	–1	2	–4
y	0	4	–0

Now, plot the points P(2, 4) and Q(–4, 0). The point B(–1, 2) has already been plotted.

Join PB and BQ and extend it on both ways.

Thus, PQ is the graph of 2x + y – 8 = 0.

The two graph lines intersect at B(–1, 2).

∴ x = –1 and y = 2