anonymous
  • anonymous
Please help.. What are the solutions to the following system of equations? y = x2 + 12x + 30 8x − y = 10 (−4, −2) and (2, 5) (−2, −4) and (2, 5) (−2, −4) and (5, 2) No Real Solutions
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
uri
  • uri
@DebbieG Help her ^_^ Uri is proud of you.
DebbieG
  • DebbieG
What methods have you learned for solving a system? Substitution, elimination?

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anonymous
  • anonymous
Elimination I think
anonymous
  • anonymous
And some substitution but I cant remember how to do that method
DebbieG
  • DebbieG
I would probably opt for substitution here, since you have a quadratic (a parabola) and a linear equation (a line). But either would work. If you solve the 2nd equation for y, then you have two equations that both give y as a function of x, right? E.g., you have y={some stuff} and y={some different stuff} so since y= both of those things, you can set them equal to each other.
DebbieG
  • DebbieG
When you do that, you'll have a single, quadratic equation to solve.
anonymous
  • anonymous
..... that makes no sense
DebbieG
  • DebbieG
What part didn't make sense? Here are your two equations: \[y=x^2+12x+30\]\[8x-y=10\]Can you solve the 2nd one for y?
anonymous
  • anonymous
Idk how.. I only know how to solve the ones with Y on the other side of the equal sign
DebbieG
  • DebbieG
You just need to rearrange 8x-y=10 so that the y is by itself. Try adding y to both sides: 8x-y+y=10+y What do you get when you simplify that? Then subtract 10 from both sides.
anonymous
  • anonymous
IM SO FREAKING LOST
anonymous
  • anonymous
THANKS GUYS. I don't need any more help
DebbieG
  • DebbieG
@jessssloukaa the idea is not to do the problem FOR her.

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