document.write( "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? \n" ); document.write( "
Algebra.Com's Answer #789243 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "This linear equation has no constant term. That means you can't find x- and y-intercepts by setting y=0 and x=0.

\n" ); document.write( "Since you can't find x- and y-intercepts, you need to use a different method for making the graph.

\n" ); document.write( "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.

\n" ); document.write( "\"3y+=+-5x\"

\n" ); document.write( "Divide by 3 to get the equation in y=mx+b form.

\n" ); document.write( "\"y+=+%28-5%2F3%29x\"
\n" ); document.write( "or
\n" ); document.write( "\"y+=+%28-5%2F3%29x%2B0\"

\n" ); document.write( "So the y-intercept (and x-intercept) is (0,0), and the slope is -5/3.

\n" ); document.write( "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).

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