Question

$$\begin{gathered} 3x+y+1=0, \\ 2x-3y+8=0. \end{gathered}$$

Answer 1

n 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)
∴ $y=\frac{2x+8}{3}$
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	2	4	0

Now, plot the points P(2, 4), 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 − 3y + 8 = 0.

The two graph lines intersect at B(−1, 2).
∴ x = −1 and y = 2