SOLUTION: I am trying to graph 3y= -5x, I tried by substituting x for 0 then y for 0. I ended up somehow getting 0,1/3. I just tried again and I was able to get 0,-3. I have looked it up to

Algebra ->  Graphs -> SOLUTION: I am trying to graph 3y= -5x, I tried by substituting x for 0 then y for 0. I ended up somehow getting 0,1/3. I just tried again and I was able to get 0,-3. I have looked it up to       Log On


   

Question 1164786: I am trying to graph 3y= -5x, I tried by substituting x for 0 then y for 0. I ended up somehow getting 0,1/3. I just tried again and I was able to get 0,-3. I have looked it up to see if i was right or not because I wasn't sure and it did something completely different. So now im all messed up and I have no idea if what I am trying is the right thing, can you help?
Found 3 solutions by josgarithmetic, MathTherapy, greenestamps:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
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I am trying to graph 3y= -5x, I tried by substituting x for 0 then y for 0. I ended up somehow getting 0,1/3. I just tried again and I was able to get 0,-3.
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Do not try to make this complicated.
If set x to 0, then 3y=-5%2A0
3y=0
y=0%2F3
y=0-----------------point contained on the line obviously the origin, (0,0).


You will have no further trouble finding the intercepts and making your graph.

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!

I am trying to graph 3y= -5x, I tried by substituting x for 0 then y for 0. I ended up somehow getting 0,1/3. I just tried again and I was able to get 0,-3. I have looked it up to see if i was right or not because I wasn't sure and it did something completely different. So now im all messed up and I have no idea if what I am trying is the right thing, can you help?
You don't need to substitute 0 for x and 0 for y, since when you solve the equation for y, b, or the y-intercept is 0, and if the y-intercept is 0, the x-intercept will also be 0 ====> matrix%282%2C3%2C+3y%2C+%22=%22%2C+-+5x%2C+y%2C+%22=%22%2C+%28-+5%2F3%29x%29
With a slope of -+5%2F3, we get  .

Starting at the y-intercept, (0, 0) and using slope %28-+5%29%2F3 ===> going DOWN 5 units, and RIGHT 3 units will lead to a 2nd point. You can now connect the points to get the graph. 

Likewise, starting at the y-intercept, (0, 0) and using slope %22+%2B+5%22%2F%28-+3%29 ===> going UP 5 units, and LEFT 3 units will lead to a 2nd point. You can now connect the points to get the graph.
That's ALL!!

Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


This linear equation has no constant term. That means you can't find x- and y-intercepts by setting y=0 and x=0.

Since you can't find x- and y-intercepts, you need to use a different method for making the graph.

Another easy way to make a graph of a linear equation is using slope-intercept form, y = mx+b. In that form, the y-intercept is (0,b) and the slope is m.

3y+=+-5x

Divide by 3 to get the equation in y=mx+b form.

y+=+%28-5%2F3%29x
or
y+=+%28-5%2F3%29x%2B0

So the y-intercept (and x-intercept) is (0,0), and the slope is -5/3.

So your first point can be (0,0). Then use the slope to find the second point. Slope = rise/run = -5/3, so, starting at (0,0), move 5 down (a rise of -5) and 3 to the right (a run of 3). Your second point is (3,-5).