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Mathematics 58 Online
OpenStudy (anonymous):

College Algebra: I'm doing 2X2 Non-Linear Systems and I'm stuck on this problem. I need to solve this system algebraically and then graph both of the equations on the same coordinate system to support my solution. 5x-y=6 y=x^2

OpenStudy (anonymous):

I fixed my question everyone!

Directrix (directrix):

5x-y=6 y=x^2 --------- In the first equation, in place of y, substitute x^2 because y = x^2.

OpenStudy (anonymous):

Alright, that's pretty much as far as I got when I started solving it :D

Directrix (directrix):

So, 5x - (x^2) = 6

OpenStudy (anonymous):

Would it be 4x^3=6?

Directrix (directrix):

5x - (x^2) = 6 -x^2 + 5x = 6 Multiply through by -1 x^2 - 5x = -6 Get all terms on one side of the equation. x^2 - 5x + 6 = 0 How does that factor? @Natasha.g.2013

OpenStudy (anonymous):

Wow

OpenStudy (anonymous):

Give me a second to factor it

Directrix (directrix):

>> 4x^3=6? No because 5x and -x^2 are not like terms.

OpenStudy (anonymous):

^right

Directrix (directrix):

@Natasha.g.2013 How did you factor this: x^2 - 5x + 6 = 0

OpenStudy (anonymous):

Sadly, I'm still trying to. I'm very slow at factoring. Takes me some time.

Directrix (directrix):

What are two numbers that multiply to 6 AND add to - 5?

OpenStudy (anonymous):

1

TheSmartOne (thesmartone):

"2 numbers"

OpenStudy (anonymous):

I got it!

OpenStudy (anonymous):

(x+1)(x-6)

OpenStudy (anonymous):

wait, never mind

TheSmartOne (thesmartone):

I'll give you a hint. It is two negative numbers.

OpenStudy (anonymous):

It's -1 and -6 huh?

TheSmartOne (thesmartone):

No... -1+(-6)=-7 so now. What other factors does 6 have besides 1 and 6 @Natasha.g.2013 @~

OpenStudy (anonymous):

Ohhh, 5 and 6

OpenStudy (anonymous):

I mean 5 and 1

TheSmartOne (thesmartone):

Does 5 and 1 multiplied equal to 6?

OpenStudy (anonymous):

crap, no

TheSmartOne (thesmartone):

And I said that the two numbers are negative

OpenStudy (anonymous):

I was thinking of addition

TheSmartOne (thesmartone):

Ok what factors does 6 have. There are only like 4.

TheSmartOne (thesmartone):

then we can take two of those factors that multiply to become 6 and add to 5 and make them negative

OpenStudy (anonymous):

Well I learned something new. I suck at simple factoring.

OpenStudy (anonymous):

2 and 3

TheSmartOne (thesmartone):

Exactly! now we just need to make both negative so that when we add them we get -5

TheSmartOne (thesmartone):

Also we could have solved it by quadratic formula or solve by squaring way..

OpenStudy (anonymous):

Alright, so it's (x-2)(x-3)=0

TheSmartOne (thesmartone):

Sorry for taking over lol @Directrix :)

TheSmartOne (thesmartone):

Yes it is. And now if we solve for x we get the y-intercepts :D

TheSmartOne (thesmartone):

Because we need to graph it.

OpenStudy (anonymous):

Wouldn't you do x-2=0 and x-3=0?

TheSmartOne (thesmartone):

Exactly :)

OpenStudy (anonymous):

x=2 and x=3

Directrix (directrix):

@TheSmartOne No worries, just help the Asker finish it.

TheSmartOne (thesmartone):

:)

TheSmartOne (thesmartone):

x^2 - 5x + 6 = 0 is in the format of Ax^2+bx+c=0 and since a>o we know that the parabola has to open upwards

TheSmartOne (thesmartone):

And we know that the y-intercepts are (3,0) and (2,0)

TheSmartOne (thesmartone):

And we can find the vertex by (-b/2a,f(-b/2a))

OpenStudy (anonymous):

So, would I plug in (2,0) and (3,0)?

TheSmartOne (thesmartone):

No we would just graph it now that we know the points.

TheSmartOne (thesmartone):

But first we should find teh vertex.

TheSmartOne (thesmartone):

the*

OpenStudy (anonymous):

Wouldn't the vertex be (6, 0) or am I completely wrong?

TheSmartOne (thesmartone):

x^2 - 5x + 6 = 0 is in the format of Ax^2+bx+c=0 So it is -b/2a to find the x-coordinate And that is not correct.

OpenStudy (anonymous):

Ohhhh

TheSmartOne (thesmartone):

So -b/2a is -(-5)/2(1) -(-5)=+5 5/2=2.5

TheSmartOne (thesmartone):

So since x=2.5 we can plug that back in and solve for x.

TheSmartOne (thesmartone):

solve for what the y-coordinate is*****

TheSmartOne (thesmartone):

So (2.5)^2 - 5(2.5) + 6 =?

OpenStudy (anonymous):

Plug 2.5 into the x^2-5x+6?

TheSmartOne (thesmartone):

yes

OpenStudy (anonymous):

I got -0.25

TheSmartOne (thesmartone):

So the vertex is (2.5,-.25)

TheSmartOne (thesmartone):

https://www.desmos.com/calculator/sr4beafkft So here is how the graph looks like :)

OpenStudy (anonymous):

So, that's it?

TheSmartOne (thesmartone):

Yes that is the graph

OpenStudy (anonymous):

Alright, so what's the answer?

OpenStudy (anonymous):

We're not done with the equation right?

TheSmartOne (thesmartone):

Yeah i guess we are done...

OpenStudy (anonymous):

Oh... According to my texbook, the answer is {(2,4),(3,9)}

TheSmartOne (thesmartone):

...

TheSmartOne (thesmartone):

y=x^2 That is what you get when you plug in x=2 and x=3 into this equation...

TheSmartOne (thesmartone):

But I do not see why we have to do that...

OpenStudy (anonymous):

I don't see where the book got 4 and 9

TheSmartOne (thesmartone):

well the y-intercepts are 2 and 3 so when you plug them in to y=x^2 you get the y value as 4 and 9 respectively so you get (2,4), (3,9)

OpenStudy (anonymous):

Oh, yeah that makes sense

OpenStudy (anonymous):

So once I got x=2 and x=3, I just had to plug it into y=x^2?

TheSmartOne (thesmartone):

yes I h=guess so

TheSmartOne (thesmartone):

I guess**

OpenStudy (anonymous):

Wow, unbelievable. Thank you so much for your help though. You must be frustrated now :D

TheSmartOne (thesmartone):

No I am not lol

TheSmartOne (thesmartone):

We are here on OS to help people :) I am also tkaing College Algebra 2 :)

TheSmartOne (thesmartone):

College Algebra too*

OpenStudy (anonymous):

I do appreciate your help though. As well as yours @Directrix. :) College Algebra is the only math course I need to take. It's not as fun and easy as it was in high school.

TheSmartOne (thesmartone):

No problem. I am in high school though lol

OpenStudy (anonymous):

Oh, well lucky you. :D

TheSmartOne (thesmartone):

But since you are new here feel free to read this :) http://openstudy.com/study#/updates/543de42fe4b0b3c6e146b5e8

OpenStudy (mathmate):

@natasha.g.2013 The x-values that you found were 2 and 3. You need to plug them into y=x^2 because you need the corresponding y-values of the two solutions in x-y couples. That's how you would get (2,4),(3,9).

OpenStudy (anonymous):

Thank you @Mathmate

OpenStudy (mathmate):

You're welcome! :)

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