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Find the points of intersection of the following lines:
x + 2y = 6
2x + 4y = -9
Solution:
Since the given equations are all first degree, then they are linear equations. They are straight lines. We can graph the two lines by getting their slope and y-intercept.
For x + 2y = 6,
x + 2y = 6
2y = - x + 6
y = -½ x + 3
slope (Δy/Δx), m = -½
y-intercept, b = 3
To trace the graph, plot 3 at the y-axis. This is your first point of the line (0, 3). Next, use the slope to get the second point. From the first point, count 2 units to the left and then 1 unit upward.
For 2x + 4y = -9,
2x + 4y = -9
4y = - 2x - 9
y = -½ x - 9/4
slope (Δy/Δx), m = -½
y-intercept, b = - 9/4
To trace the graph, plot - 9/4 at the y-axis. This is your first point of the line (0, - 9/4). Next, use the slope to get the second point. From the first point, count 2 units to the left and then 1 unit upward.
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From the graph, the two lines are parallel because their slopes are the same which is -½. The two lines will never meet how far they are extended. When you solve for x and y from the two given equations, their x and y will be equal to zero. From the two given equations,
x + 2y = 6
2x + 4y = -9
Multiply the first equation by 2 and -1 at the second equation. Add the two equations and let's see what will happen to x and y.
2 (x + 2y = 6) 2x + 4y = 12
→
- 1 (2x + 4y = -9) -2x - 4y = 9
______________
0 ≠ 21
Since their x and y are equal to zero, then we cannot solve for x and y. Also, the right side of the final equation is not zero. Therefore, the two lines are parallel to each other.