Question 142966
3x-1=5x+y
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This is a linear equation, what we need to do first is put it in slope-intercept form: y=mx+b.
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For this equation its really simple. All we do is subtract the 5x off from both sides, which will look like this:
3x-1-5x=5x-5x+y....add like terms and we have...
-2x-1=y
This will be the form of the equation that we graph.
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What will really help you when graphing linear lines is knowing the components of the equation. 
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I always look at the y-intercept first:
Or in otherwords the B
In our case its the -1
This is the point where our line will cross the y-axis, which is the one going up in down.
For us we make our point at -1 on the y-axis (our points is 0,-1)
We can also find this the longer way by throwing in a 0 from x in our equation and solving for y, which will end up being -1 anyways. 
{{{drawing(200,200,-4,4,-4,4,
grid(1),
circle(0,-1,0.12))}}}

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Then I look at m, or the number before x,which is our slope.
All slope is is how slanted our line will be.
In our case we have the number -2 (just to remind us of the equation -2x-1=y)
Any number can be written with a one under it, so we have -2/1 as our slope. That means we go down 2 from our first point(the b, or the y-intercept), and over 1 to the right. 
{{{drawing(200,200,-4,4,-4,4,
grid(1),
circle(0,-1,0.12),
circle(1,-3,0.12))}}}
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We have to be careful, though, and make sure that our slope is negative when we graph it, in otherwords going down to the right
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With the two points we just found we have enough to graph our line. All we do is draw a line going through both of our points.
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our graph will end up looking like this when we're finished:

{{{drawing(200,200,-4,4,-4,4,
grid(1),
circle(0,-1,0.12),
circle(1,-3,0.12),
red(line(-2.5,4,1.5,-4)))}}}