SOLUTION: Graph the linear equations 8x+y=0

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Question 116720: Graph the linear equations
8x+y=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Graphing Linear Equations


8%2Ax%2B1%2Ay=0Start with the given equation



1%2Ay=0-8%2Ax Subtract 8%2Ax from both sides

y=%281%29%280-8%2Ax%29 Multiply both sides by 1

y=%281%29%280%29-%281%29%288%29x%29 Distribute 1

y=0-%288%29x Multiply

y=-8%2Ax%2B0 Rearrange the terms

y=-8%2Ax%2B0 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=-8 (the slope) and b=0 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-1

y=-8%2A%28-1%29%2B0

y=8%2B0 Multiply

y=8 Add

So here's one point (-1,8)





Now lets find another point

Plug in x=0

y=-8%2A%280%29%2B0

y=0%2B0 Multiply

y=0 Add

So here's another point (0,0). Add this to our graph





Now draw a line through these points

So this is the graph of y=-8%2Ax%2B0 through the points (-1,8) and (0,0)


So from the graph we can see that the slope is -8%2F1 (which tells us that in order to go from point to point we have to start at one point and go down -8 units and to the right 1 units to get to the next point), the y-intercept is (0,0)and the x-intercept is (0,0) . So all of this information verifies our graph.


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


So we have one point (0,0)






Now since the slope is -8%2F1, this means that in order to go from point to point we can use the slope to do so. So starting at (0,0), we can go down 8 units


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



Now draw a line through those points to graph y=-8%2Ax%2B0


So this is the graph of y=-8%2Ax%2B0 through the points (0,0) and (1,-8)