Question 121541
I have tried to figure out how to properly graph this problem however, when I looked at the answer already provided and what I had come up with became confused because they did not seem to match. x-2y=8 and 3x-2y=12. Here is what I have worked out so far.<br>

x-2y = 8
-2y = -x + 8
y = x/2 - 4
y intercept: (0, -4)
slope: 1/2<br>

3x-2y = 12
-2y = -3x + 12
y = 3x/2 - 6
y intercept: (0, -6)
slope: 3/2<br>

I assume I would graph with the following points (0,-4) and (4,0) for the second one I assumed I would graph the following points (0,-6) (6,0).
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well your assumptions are what is stopping you from this, because the second set of points aren't on the graphs but up to that point you were right on.<br>

To find the second set of points to graph you can either set both equations equal to 0 and solve(finding the x-intercepts) or just move your slope away from  your origonal points.  <br>

Method 1- finding your x-intercepts and then graphing those.<br>

0 = x/2 - 4 --> add 4 to both sides
4 = x/2 --> multiply both sides by 2
8 = x<br>

0 = 3x/2 - 6 --> add 6 to both sides
6 = 3x/2 --> multiply both sides by 2
12 = 3x --> divide both sides by 3
4 = x<br> 

so the first equation will cross the x-axis at (8,0) and the second equation will cross the y-axis at (4,0). Now you have two points for each line so you can graph them and see where they cross.<br>

Method 2- using the slope<br>

the first y-intercept is (0,-4) and the slope of that line is 1/2 but remember slope is rise over run so all you need to do is add 1 to the y value and 2 to the x value and you have a second point and can graph that line.  the new point is (2,-3).<br>

the second y-intercept is (0,-6) and the slope of that line is 3/2 but again remember slope is rise over run so all you need to do is add 3 to the y value and 2 to the x value and you have a second point and can graph that line.  the new point is (2,-3)