Graphing Linear Equations

[travis+]

I'm experiencing a bit of confusion remembering how to graph basic lines.

I know that I can find x and y intercepts because the equation for the x-axis is that y=0 and the equation for the y-axis is that x=0.

So, for example, x+y=2
x + (0) = 2, therefore the x-intercept is (2,0)
and (0) + y = 2, therefore the y-intercept is (0,2)
So I would mark each of those points on on the graph, and draw a line between them, as seen in the picture below.



I'm having trouble with the second problem.

2x+3y = 0

I would assume that I am again looking for the x and y intercepts, so I put in
2(0) + 3y = 0
3y = 0
y = 0

And basically the same thing, for x... the intercept for each would be (0,0)... which suggests to me that there is not really a line here, probably just a point on the map, or maybe a line going out in all directions, I don't know. But the graph on the answer key (above) shows a line intersecting through (0,0) and (3, -2). Can you explain how I am supposed to get to that point?

Also, just jumping ahead to the third problem for a second. x+4=0 is the equation being offered. I can see that x=-4, but what do I do with y in this case? My gut would have been to say that this is a horizontal line, but the answer shows it as a vertical line. Does this mean that when y or x is not present in the equation, any number on the number line is a solution?

Thanks!

 

[wjm11]

I'm experiencing a bit of confusion remembering how to graph basic lines.

I know that I can find x and y intercepts because the equation for the x-axis is that y=0 and the equation for the y-axis is that x=0.

So, for example, x+y=2
x + (0) = 2, therefore the x-intercept is (2,0)
and (0) + y = 2, therefore the y-intercept is (0,2)
So I would mark each of those points on on the graph, and draw a line between them, as seen in the picture below.

View attachment 2736

I'm having trouble with the second problem.

2x+3y = 0

I would assume that I am again looking for the x and y intercepts, so I put in
2(0) + 3y = 0
3y = 0
y = 0

And basically the same thing, for x... the intercept for each would be (0,0)... which suggests to me that there is not really a line here, probably just a point on the map, or maybe a line going out in all directions, I don't know. But the graph on the answer key (above) shows a line intersecting through (0,0) and (3, -2). Can you explain how I am supposed to get to that point?

Also, just jumping ahead to the third problem for a second. x+4=0 is the equation being offered. I can see that x=-4, but what do I do with y in this case? My gut would have been to say that this is a horizontal line, but the answer shows it as a vertical line. Does this mean that when y or x is not present in the equation, any number on the number line is a solution?

Thanks!

Your approach of finding the x and y intercepts is fine, except when the line passes through the origin. Then you only have one point (with your approach), and you need two points to define the line graphically. I suggest you start here: http://www.purplemath.com/modules/graphlin4.htm#top
for a good start on graphing linear equations. There are four pages of examples and explanations that will answer the questions you've posed thus far.