SOLUTION: 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
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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.
x-2y = 8
-2y = -x + 8
y = x/2 - 4
y intercept: (0, -4)
slope: 1/2
3x-2y = 12
-2y = -3x + 12
y = 3x/2 - 6
y intercept: (0, -6)
slope: 3/2
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). I think I have confused the actual points in which I am to graph. Could you please tell me if I am graphing the correct points? I do not have the ISBN for this book becasue it is an online class. Thanks and have a great day!!
Found 2 solutions by Fombitz, algebrapro18:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
The slope-intercept form is one way of doing it but I don't think it's as intuitive as finding the x and y intercepts.
Let's do it a different way.
Find the x-intercept of the graph.
Then find the y-intercept of the graph.
When y=0, then
When x=0, then
Now you have two points on the graph (0,-4) and (8,0).
It’s easier to graph this way than with a point and the slope, I think.
First graph the two points.
Then draw the line containing the two points.
When y=0, then
When x=0, then
The two points on the graph are(4,0) and (0,6).
Graph the two points.
Then draw the line containing the two points.
You can then look at your graphs and find the slope by using change in y divided by change in x.
Answer by algebrapro18(249) (Show Source): You can put this solution on YOUR website!
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.
x-2y = 8
-2y = -x + 8
y = x/2 - 4
y intercept: (0, -4)
slope: 1/2
3x-2y = 12
-2y = -3x + 12
y = 3x/2 - 6
y intercept: (0, -6)
slope: 3/2
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).
--------------------------------------------------------------------------------
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.
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.
Method 1- finding your x-intercepts and then graphing those.
0 = x/2 - 4 --> add 4 to both sides
4 = x/2 --> multiply both sides by 2
8 = x
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
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.
Method 2- using the slope
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).
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)
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