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Or anyone that can help me

what do you mean by "solve?"

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I'm not sure thats all they told us.

Okay thank you

Do I solve for x?

do the substitution method where you multiple one of the equations by -1

Okay I will. For the substitution method do I make y -1? Sorry I'm not that great at math.

Wow! Thank you so much!

|dw:1433532657818:dw|

I'm stuck at this point. Did I do it wrong?

\[x^2+3=x+5\]
collect everything on one side, what do you get?

0 = -x^2 +x + 2

Yes sorry my dad had to use the computer

I know how to solve using quadratic formula

|dw:1433535440928:dw|

did i do that right?

I think you probably just didn't read your handwriting...

Yeah sorry it was really messy.

correct answer is...

My solutions then would be 2 and -2?

let's try them out!
\[x^2-x-2=0\]\[(2)^2-(2)-2=0\]\[4-2-2=0\checkmark\]
you do the other one (x=-2)

we are just verifying that we solved for \(x\) correctly here...

\[x^2-x-2 = 0\]
\[-2^2-(-2)-2 =0\]
4+2-2 = 0
4 = 0

Hmm. maybe one of those solutions is incorrect :-)

-2 won't be a solution

Oh, sorry.

Thanks a million!

okay so then y will equal 7 and -4

(2,7) (-1, -4)

Yep! And if you plug them all in to all the equations, they should all work.

Great! Thank you :)

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