SOLUTION: dealing with solving linear systems how do i do y-2x=0 6y+2x=o

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Question 397539: dealing with solving linear systems how do i do y-2x=0 6y+2x=o
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!


y-2x=0
+6y%2B2x=o
or
-2x+%2B+y+=+0
x+%2B+6y+=+0

Solved by pluggable solver: Solve the System of Equations by Graphing



Start with the given system of equations:


-2x%2By=0

1x%2B6y=0





In order to graph these equations, we need to solve for y for each equation.




So let's solve for y on the first equation


-2x%2By=0 Start with the given equation



1y=0%2B2x Add 2+x to both sides



1y=%2B2x%2B0 Rearrange the equation



y=%28%2B2x%2B0%29%2F%281%29 Divide both sides by 1



y=%28%2B2%2F1%29x%2B%280%29%2F%281%29 Break up the fraction



y=2x%2B0 Reduce



Now lets graph y=2x%2B0 (note: if you need help with graphing, check out this solver)



+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+2x%2B0%29+ Graph of y=2x%2B0




So let's solve for y on the second equation


1x%2B6y=0 Start with the given equation



6y=0-x Subtract +x from both sides



6y=-x%2B0 Rearrange the equation



y=%28-x%2B0%29%2F%286%29 Divide both sides by 6



y=%28-1%2F6%29x%2B%280%29%2F%286%29 Break up the fraction



y=%28-1%2F6%29x%2B0 Reduce





Now lets add the graph of y=%28-1%2F6%29x%2B0 to our first plot to get:


+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+2x%2B0%2C%28-1%2F6%29x%2B0%29+ Graph of y=2x%2B0(red) and y=%28-1%2F6%29x%2B0(green)


From the graph, we can see that the two lines intersect at the point (0,0) (note: you might have to adjust the window to see the intersection)