SOLUTION: How would you graph 2x-y=-2? Or get it into y=mx+b?

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Question 553116: How would you graph 2x-y=-2? Or get it into y=mx+b?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Graphing Linear Equations


2%2Ax-1%2Ay=-2Start with the given equation



-1%2Ay=-2-2%2Ax Subtract 2%2Ax from both sides

y=%28-1%29%28-2-2%2Ax%29 Multiply both sides by -1

y=%28-1%29%28-2%29%2B%281%29%282%29x%29 Distribute -1

y=2%2B%282%29x Multiply

y=2%2Ax%2B2 Rearrange the terms

y=2%2Ax%2B2 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=2 (the slope) and b=2 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-5

y=2%2A%28-5%29%2B2

y=-10%2B2 Multiply

y=-8 Add

So here's one point (-5,-8)





Now lets find another point

Plug in x=-4

y=2%2A%28-4%29%2B2

y=-8%2B2 Multiply

y=-6 Add

So here's another point (-4,-6). Add this to our graph





Now draw a line through these points

So this is the graph of y=2%2Ax%2B2 through the points (-5,-8) and (-4,-6)


So from the graph we can see that the slope is 2%2F1 (which tells us that in order to go from point to point we have to start at one point and go up 2 units and to the right 1 units to get to the next point) the y-intercept is (0,2)and the x-intercept is (-1,0) . So all of this information verifies our graph.


We could graph this equation another way. Since b=2 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,2).


So we have one point (0,2)






Now since the slope is 2%2F1, this means that in order to go from point to point we can use the slope to do so. So starting at (0,2), we can go up 2 units


and to the right 1 units to get to our next point



Now draw a line through those points to graph y=2%2Ax%2B2


So this is the graph of y=2%2Ax%2B2 through the points (0,2) and (1,4)