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Find the points of intersection of the following lines:
x + 2y - 3 = 0
2x - 3y + 8 = 0
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 - 3 = 0,
x + 2y - 3 = 0
2y = -x + 3
y = - ½ x + 3/2
slope (∆y/∆x), m = - ½
y-intercept = 3/2
To trace the graph, plot 3/2 at the y-axis. This is your first point of the line (0, 3/2). 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 - 3y + 8 = 0,
2x - 3y + 8 = 0
3y = 2x + 8
y = ⅔ x + 8/3
slope (∆y/∆x), m = ⅔
y-intercept = 8/3 or 2 ⅔
To trace the graph, plot 8/3 at the y-axis. This is your first point of the line (0, 8/3). Next, use the slope to get the second point. From the first point, count three units to the right and then 2 units upward.
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| Photo by Math Principles in Everyday Life |
From the graph, their point of intersection is (-1, 2). To get their actual point of intersection, we have to use the two given equations and solve for x and y, we have
x + 2y - 3 = 0
2x - 3y + 8 = 0
Multiply the 1st equation by 2 and -1 at the 2nd equation. Add the two equations in order to eliminate x and solve for the value of y.
2(x + 2y - 3 = 0) 2x + 4y - 6 = 0
→
-1(2x - 3y + 8 = 0) -2x + 3y - 8 = 0
________________
7y - 14 = 0
7y = 14
y = 2
Substitute y to either of the two equations,
x + 2y - 3 = 0
x + 2(2) - 3 = 0
x + 4 - 3 = 0
x + 1 = 0
x = -1
Therefore, their point of intersection is P(-1, 2).
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| Photo by Math Principles in Everyday Life |