Question

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

Answer 1

I fixed my question everyone!

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Would it be 4x^3=6?

Answer 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

Answer 7

Wow

Answer 8

Give me a second to factor it

Answer 9

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

Answer 10

^right

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

1

Answer 15

"2 numbers"

Answer 16

I got it!

Answer 17

(x+1)(x-6)

Answer 18

wait, never mind

Answer 19

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

Answer 20

It's -1 and -6 huh?

Answer 21

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

Answer 22

Ohhh, 5 and 6

Answer 23

I mean 5 and 1

Answer 24

Does 5 and 1 multiplied equal to 6?

Answer 25

crap, no

Answer 26

And I said that the two numbers are negative

Answer 27

I was thinking of addition

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

2 and 3

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

Sorry for taking over lol @Directrix :)

Answer 36

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

Answer 37

Because we need to graph it.

Answer 38

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

Answer 39

Exactly :)

Answer 40

x=2 and x=3

Answer 41

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

Answer 42

:)

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

But first we should find teh vertex.

Answer 49

the*

Answer 50

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

Answer 51

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.

Answer 52

Ohhhh

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

yes

Answer 59

I got -0.25

Answer 60

So the vertex is (2.5,-.25)

Answer 61

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

Answer 62

So, that's it?

Answer 63

Yes that is the graph

Answer 64

Alright, so what's the answer?

Answer 65

We're not done with the equation right?

Answer 66

Yeah i guess we are done...

Answer 67

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

Answer 68

...

Answer 69

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

Answer 70

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

Answer 71

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

Answer 72

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)

Answer 73

Oh, yeah that makes sense

Answer 74

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

Answer 75

yes I h=guess so

Answer 76

I guess**

Answer 77

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

Answer 78

No I am not lol

Answer 79

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

Answer 80

College Algebra too*

Answer 81

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.

Answer 82

No problem. I am in high school though lol

Answer 83

Oh, well lucky you. :D

Answer 84

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

Answer 85

@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).

Answer 86

Thank you @Mathmate

Answer 87

You're welcome! :)