SOLUTION: how to solve 2x+y=0 to graph isbn 0321123388-3

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Question 83601This question is from textbook
: how to solve 2x+y=0 to graph isbn 0321123388-3This question is from textbook

Answer by jim_thompson5910(21667) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Graphing Linear Equations


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



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

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

y=%281%29%280%29-%281%29%282%29x%29 Distribute 1

y=0-%282%29x Multiply

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

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

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

So to graph this equation lets plug in some points

Plug in x=-4

y=-2%2A%28-4%29%2B0

y=8%2B0 Multiply

y=8 Add

So here's one point (-4,8)



drawing%28+600%2C+600%2C+-14%2C7%2C+-4%2C+18%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+-4%2C8%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-4%2C8%2C+.1%2C+1.5+%29+%29%0D%0A++%29

Now lets find another point

Plug in x=-3

y=-2%2A%28-3%29%2B0

y=6%2B0 Multiply

y=6 Add

So here's another point (-3,6). Add this to our graph


drawing%28+600%2C+600%2C+-14%2C7%2C+-4%2C+18%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+-3%2C6%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-3%2C6%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-4%2C8%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-4%2C8%2C+.1%2C+1.5+%29+%29%0D%0A++%29


Now draw a line through these points

drawing%28+600%2C+600%2C+-14%2C7%2C+-4%2C+18%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-14-10%2C-2%2A%28-14-10%29%2B0%2C7%2B10%2C-2%2A%287%2B10%29%2B0%29%2C%0D%0A++++blue%28+circle%28+-3%2C6%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-3%2C6%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-4%2C8%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-4%2C8%2C+.1%2C+1.5+%29+%29%0D%0A++%29 So this is the graph of y=-2%2Ax%2B0 through the points (-4,8) and (-3,6)


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 down -2 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)


drawing%28+600%2C+600%2C+-10%2C11%2C+-12%2C+10%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++%0D%0A++++blue%28+circle%28+0%2C0%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C0%2C+.1%2C+1.5+%29+%29%0D%0A++%29



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 (0,0), we can go down 2 units

drawing%28++600%2C+600%2C+-10%2C11%2C+-12%2C+10%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+0%2C0%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C0%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+0%2B%28-2%2F2%29%2C+2%2C+-2%2C+90%2C+270+%29+%29%0D%0A++%0D%0A++%29
and to the right 1 units to get to our next point

drawing%28++600%2C+600%2C+-10%2C11%2C+-12%2C+10%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+0%2C0%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C0%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+1%2C-2%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+1%2C-2%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+0%2B%28-2%2F2%29%2C+2%2C+-2%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%281%2F2%29%2C+-2%2C+1%2C+2%2C+0%2C+180+%29+%29%0D%0A++%0D%0A++%29

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


drawing%28+600%2C+600%2C+-10%2C11%2C+-12%2C+10%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-10-10%2C-2%2A%28-10-10%29%2B0%2C11%2B10%2C-2%2A%2811%2B10%29%2B0%29%2C%0D%0A++++blue%28+circle%28+1%2C-2%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+1%2C-2%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+0%2C0%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C0%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+0%2B%28-2%2F2%29%2C+2%2C+-2%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%281%2F2%29%2C+-2%2C+1%2C+2%2C+0%2C+180+%29+%29%0D%0A++%29 So this is the graph of y=-2%2Ax%2B0 through the points (0,0) and (1,-2)