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- 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

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- anonymous

- jamiebookeater

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- anonymous

- uri

@DebbieG Help her ^_^ Uri is proud of you.

- DebbieG

What methods have you learned for solving a system? Substitution, elimination?

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- anonymous

Elimination I think

- anonymous

And some substitution but I cant remember how to do that method

- 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

When you do that, you'll have a single, quadratic equation to solve.

- anonymous

..... that makes no sense

- 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

Idk how.. I only know how to solve the ones with Y on the other side of the equal sign

- 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

IM SO FREAKING LOST

- anonymous

THANKS GUYS. I don't need any more help

- DebbieG

@jessssloukaa the idea is not to do the problem FOR her.

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