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
I fixed my question everyone!
5x-y=6 y=x^2 --------- In the first equation, in place of y, substitute x^2 because y = x^2.
Alright, that's pretty much as far as I got when I started solving it :D
So, 5x - (x^2) = 6
Would it be 4x^3=6?
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
Wow
Give me a second to factor it
>> 4x^3=6? No because 5x and -x^2 are not like terms.
^right
@Natasha.g.2013 How did you factor this: x^2 - 5x + 6 = 0
Sadly, I'm still trying to. I'm very slow at factoring. Takes me some time.
What are two numbers that multiply to 6 AND add to - 5?
1
"2 numbers"
I got it!
(x+1)(x-6)
wait, never mind
I'll give you a hint. It is two negative numbers.
It's -1 and -6 huh?
No... -1+(-6)=-7 so now. What other factors does 6 have besides 1 and 6 @Natasha.g.2013 @~
Ohhh, 5 and 6
I mean 5 and 1
Does 5 and 1 multiplied equal to 6?
crap, no
And I said that the two numbers are negative
I was thinking of addition
Ok what factors does 6 have. There are only like 4.
then we can take two of those factors that multiply to become 6 and add to 5 and make them negative
Well I learned something new. I suck at simple factoring.
2 and 3
Exactly! now we just need to make both negative so that when we add them we get -5
Also we could have solved it by quadratic formula or solve by squaring way..
Alright, so it's (x-2)(x-3)=0
Sorry for taking over lol @Directrix :)
Yes it is. And now if we solve for x we get the y-intercepts :D
Because we need to graph it.
Wouldn't you do x-2=0 and x-3=0?
Exactly :)
x=2 and x=3
@TheSmartOne No worries, just help the Asker finish it.
:)
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
And we know that the y-intercepts are (3,0) and (2,0)
And we can find the vertex by (-b/2a,f(-b/2a))
So, would I plug in (2,0) and (3,0)?
No we would just graph it now that we know the points.
But first we should find teh vertex.
the*
Wouldn't the vertex be (6, 0) or am I completely wrong?
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.
Ohhhh
So -b/2a is -(-5)/2(1) -(-5)=+5 5/2=2.5
So since x=2.5 we can plug that back in and solve for x.
solve for what the y-coordinate is*****
So (2.5)^2 - 5(2.5) + 6 =?
Plug 2.5 into the x^2-5x+6?
yes
I got -0.25
So the vertex is (2.5,-.25)
https://www.desmos.com/calculator/sr4beafkft So here is how the graph looks like :)
So, that's it?
Yes that is the graph
Alright, so what's the answer?
We're not done with the equation right?
Yeah i guess we are done...
Oh... According to my texbook, the answer is {(2,4),(3,9)}
...
y=x^2 That is what you get when you plug in x=2 and x=3 into this equation...
But I do not see why we have to do that...
I don't see where the book got 4 and 9
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)
Oh, yeah that makes sense
So once I got x=2 and x=3, I just had to plug it into y=x^2?
yes I h=guess so
I guess**
Wow, unbelievable. Thank you so much for your help though. You must be frustrated now :D
No I am not lol
We are here on OS to help people :) I am also tkaing College Algebra 2 :)
College Algebra too*
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.
No problem. I am in high school though lol
Oh, well lucky you. :D
But since you are new here feel free to read this :) http://openstudy.com/study#/updates/543de42fe4b0b3c6e146b5e8
@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).
Thank you @Mathmate
You're welcome! :)
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