anonymous
  • anonymous
how do i solve this system? (attached equation)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1378612862871:dw|
Psymon
  • Psymon
You have to do substitution. You would take the bottom equation. x + y = 4, and solve for either x or y. So you could have x = 4-y or y = 4-x, your choice. Then take that and substitute it into the 1st equation and solve for the variable.
anonymous
  • anonymous
@Psymon yeah I did that I got x=4-y and substituted it and got 7y^2-55=0 which looks wrong

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Psymon
  • Psymon
Lets see: \[7y ^{2}+(4-y)^{2}=64\] \[7y ^{2}+16-8y+y ^{2}=64\] \[8y ^{2}-8y-48 = 0 \implies 8(y ^{2}-y - 6) = 0\] Now you would factor that and solve for y
anonymous
  • anonymous
y=3 and y= -2?
Psymon
  • Psymon
Yep, exactly. So now test both points into one of the two equations and solve for x.
anonymous
  • anonymous
why doesn't it matter which equation to substitute it in?
Psymon
  • Psymon
Because a solution has to work for both equations. A solution to a system of equations is a point where both equations intersect on a graph. That means both graphs must share the same point. If they share the same point, then the x value with produce the same y-value in both equations if its a solution. Basically, if the values for y we get are true, youll get the same x no matter which equation you put it in to.
anonymous
  • anonymous
so I got (1,3) and (6,-2)
Psymon
  • Psymon
Yep, thatd be right. So those are the two points where the graphs intersect: |dw:1378614925168:dw|
anonymous
  • anonymous
thank you very much Psymon!
Psymon
  • Psymon
Yeah, np :3

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