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प्रश्न
Solve the system of equations graphically:
3x + y + 1 = 0,
2x - 3y + 8 = 0
उत्तर
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
