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Mathematics 23 Online
OpenStudy (anonymous):

use substition to find x and y (a) x-y=-8 (b) x^2-y=-2

OpenStudy (kayders1997):

Hi

OpenStudy (kayders1997):

So these equations must go together, what would you like to solve for x or y is doesn't matter which one we do

OpenStudy (kayders1997):

Do you know how to solve for x or y?

OpenStudy (anonymous):

yeah I know how to solve, I was just wondering how to get rid of the x^2 on the second equation

OpenStudy (anonymous):

I rearranged the first equation so it looked lie x=-y-8, then I plugged it into the second equation

OpenStudy (kayders1997):

You don't have to really red rid of it if these are a system of equations

OpenStudy (kayders1997):

Or are these separate?

OpenStudy (anonymous):

(-y-8)^2 -y=-2 This is what I end up with

OpenStudy (kayders1997):

You have to use the foil method because remember that (-y-8)^2 means (-y-8)(-y-8)

OpenStudy (kayders1997):

If something is squared like that it will always be the inside times the inside

OpenStudy (kayders1997):

Does this make sense?

OpenStudy (anonymous):

so it will end up looking like y^2 +8y+8y+64-y=-2

OpenStudy (kayders1997):

Just a second

OpenStudy (kayders1997):

I caught something when you moved the y to the other side

OpenStudy (anonymous):

It's suppose to be a + not a -

OpenStudy (kayders1997):

You have to do the opposite so it's subracting y in the original equation so your going to have to add by y so it will be a positive sign

OpenStudy (kayders1997):

Yes

OpenStudy (anonymous):

y^2 +8y+8y+64+y=-2 then combine like terms and you get y^2+17y =-66 (is this right)

OpenStudy (kayders1997):

Yes

OpenStudy (anonymous):

ok, this is where I got

OpenStudy (kayders1997):

I would keep the 66 on the other side though so you can factor

OpenStudy (anonymous):

...stuck (sorry hit enter)

OpenStudy (anonymous):

would 11 and 6 work

OpenStudy (kayders1997):

yes

OpenStudy (kayders1997):

So than y would equal?

OpenStudy (anonymous):

would y have two answers

OpenStudy (kayders1997):

Yes but 11 and 6 aren't it because you have (y+11)(y+6)=0

OpenStudy (anonymous):

I know we would have to set both of those factors = 0 so it's actually -11 and -6

OpenStudy (kayders1997):

Yes good

OpenStudy (anonymous):

ok, so which one do we use to plug back into our equation to get x

OpenStudy (kayders1997):

It doesn't matter they would both be the same answers but since the first one is easiest I would you the first equation

OpenStudy (anonymous):

so if I were to graph this, there should be two spots on the graphs where they touch because we have two sets of ordered pairs right

OpenStudy (kayders1997):

Hmmmmmmmm

OpenStudy (anonymous):

just checked it! And it does have two points

OpenStudy (kayders1997):

Omg I'm sorry I missed something again It would be easier If I would have done this on paper x=y-8 so you will have a -16y not positive I am so sorry

OpenStudy (kayders1997):

Because that that would be (y-8)^2 y^2-16x+64-y+2=0

OpenStudy (kayders1997):

So y^2-17y+66=0

OpenStudy (anonymous):

that's okay, I figured that out as well

OpenStudy (kayders1997):

I don't know why I missed two errors but now you would just get positive y values not negative and than get your corresponding x values

OpenStudy (kayders1997):

So we were close the first time

OpenStudy (kayders1997):

Does that make sense?

OpenStudy (anonymous):

yep, I figured that out, as soon as I graphed the two equations! Thanks for the help

OpenStudy (kayders1997):

No problem :)

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