Help Please!!!
Solve the following system of equations and show all work.
y = x2 + 3
y = x + 5

- anonymous

Help Please!!!
Solve the following system of equations and show all work.
y = x2 + 3
y = x + 5

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- jamiebookeater

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- anonymous

@mathmale

- anonymous

Or anyone that can help me

- anonymous

what do you mean by "solve?"

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## More answers

- anonymous

I'm not sure thats all they told us.

- anonymous

you would write it: x+5=x2+3 and then solve from there. the reason that is corrret is because youre just repacing y with what y is equal to- y

- anonymous

Okay thank you

- anonymous

Do I solve for x?

- anonymous

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

- NotTim

Do not do what ew.sorry said. it may work, but @arek 's method is much, much safer.

- anonymous

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

- whpalmer4

\[y=x^2+3\]\[y=x+5\]
Those are two equations to solve.
A couple of routes you can take:
substitution
setting the two right hand sides equal to each other
substitution means solving one equation for one variable in terms of the other. Then you substitute for that variable in the other equation. In this problem, this would a fine way to go because the first step has already been done for you: you have \(y=x+5\) as one of the equations, and that gives you \(y\) in terms of \(x\). Take the other equation \((y = x^2+3)\) and replace \(y\) with \(x+5\), then solve for the values of \(x\). There will be two solutions thanks to the \(x^2\) term, although in some cases they may turn out to be the same value.
another approach (which is really substitution as well) is to notice that we have two equations with the same thing on one side. Here, both equations have \(y\) by itself on one side of the equals sign. We can simply set the other two sides equal to each other, because they are equal to the same thing. After that, you solve just as before.
In both approaches, after you find the value(s) of \(x\), you plug them into one of the original equations to find the corresponding values of \(y\).

- whpalmer4

Another approach would be to graph the two equations and observe where they intersect. \[y=x^2+3\]is a parabola, and \[y=x+5\]is a line. The line crosses the parabola in two locations, which are the solutions to this system.

- whpalmer4

And yet another approach, if you had the luxury of a multiple-choice problem where you could know that the answers are provided and you merely need to identify the correct answers, would be to plug each set of answers into all of the equations and select the answer choice where all of the equations work. Note well that you can have "answers" that work for some equations but all...the technical term for them is "wrong" :-)

- anonymous

Wow! Thank you so much!

- anonymous

|dw:1433532657818:dw|

- anonymous

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

- whpalmer4

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

- anonymous

0 = -x^2 +x + 2

- whpalmer4

Yes, or \[x^2-x-2 = 0 \]
Do you know how to factor that equation, or solve it with the quadratic formula, or complete the square?

- whpalmer4

@torvia you still here?

- anonymous

Yes sorry my dad had to use the computer

- anonymous

I know how to solve using quadratic formula

- anonymous

|dw:1433535440928:dw|

- anonymous

did i do that right?

- whpalmer4

close, but not close enough. I'm intrigued by how you went from \[\sqrt{1+8}\] to \[\sqrt{4}\]in particular :-)

- whpalmer4

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

- anonymous

Yeah sorry it was really messy.

- whpalmer4

correct answer is...

- anonymous

My solutions then would be 2 and -2?

- whpalmer4

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)

- whpalmer4

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

- anonymous

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

- whpalmer4

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

- anonymous

-2 won't be a solution

- whpalmer4

by the way, you really need to write \[(-2)^2\]instead of \[-2^2\]
The first one is \((-2)*(-2) = 4\) and the second one is \(-(2)(2) = -4\)

- anonymous

Oh, sorry.

- whpalmer4

I'll speed things along a little bit:
\[\frac{1\pm\sqrt{9}}{2} = \frac{1+3}{2},\frac{1-3}{2} = \frac{4}{2},\frac{-2}{2} = 2,-1\]

- anonymous

Thanks a million!

- whpalmer4

and if we try \(-1\) in the equation:
\[(-1)^2 -(-1)-2=0\]\[1+1-2=0\checkmark\]
Now, you can plug those two values of \(x\) into one of the equations and find the corresponding values of \(y\).

- anonymous

okay so then y will equal 7 and -4

- anonymous

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

- whpalmer4

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

- anonymous

Great! Thank you :)

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