Question

Solve the system Equation... Attempt 2 ;-;

Answer 1

y = -x + 7 and y = 0.5(x - 3)^2

Answer 2

0.5(x - 3)^2 = -x + 7?

Answer 3

Hint: Solve this for x;

y = -x + 7

But it doesn't matter which way you substitute, you're going to end up with a quadratic expression either way you do it. I'd at least multiply the second equation by 2 on both sides to get rid of that decimal.

Answer 4

so Multiply 0.5 and -x by 2 @Hero?

Answer 5

I'm Lost big time.

Answer 6

2(y = -x + 7) and 2(y = 0.5(x - 3)^2)

Answer 7

Oh.

Answer 8

So it's y = -2x + 14 for the first equation?

Answer 9

More like 2y = -2x + 14

Answer 10

Oh.

Answer 11

Anyway, Not sure how useful it is to multiply both equations by 2. Just trying to get rid of that decimal. I personally don't like decimals.

Answer 12

Oh, me too.

Answer 13

My Teacher said this to me when I asked for help, and I don't quite get it.

Great, now Multiply both sides by 2 and then FOIL: -2x + 14 = x^2 - 6x + 9 Simplify: x^2 - 4x - 5 = 0 then factor to figure out the x's, then sub it x's to get y. You should end with two coordinates (x,y) where the graph's intersect.

Answer 14

Yep, that looks good actually.

Answer 15

Oh, so now what?

Answer 16

Well, you have to actually factor x^2 - 4x - 5

Answer 17

Oh

Answer 18

i'm so sorry, im lost right now. My Brain is Blank....

Answer 19

Rip my brian. ;-;

Answer 20

Do you know how to factor quadratic expressions?

Answer 21

Uhh, Yeah... I think. I might just need a reminder.

Answer 22

... a "refresher" you mean.

Answer 23

Yeah.. ;-;

Answer 24

We need to factor \(x^2 - 4x - 5\) so a = 1, b = -4, and c = -5

And we need to find two numbers \(m\) and \(n\) such that
\(m + n = |b|\)
\(m \times n = ac\)

Answer 25

Oh, Okay..

Answer 26

Basically \(ac = -5\), \(|b| = 4\)
and
\(x^2 - 4x - 5\) can be re-written as \(x^2 -(5 - 1)x - 5\)

Answer 27

Oh

Answer 28

Then continue to factor by grouping
\(x^2 - 5x + x - 5\)
\(x(x - 5) + 1(x - 5)\)
\((x - 5)(x + 1)\)

Set each factor to zero then solve for x.

Plug each x back into the original equation and solve for y.

Answer 29

Oh...

Answer 30

OH, I see

Answer 31

Wati, so Just do FOIL on (x - 5)(x - 1)?

Answer 32

What I meant to do is ask, Is that the correct expression? And no we don't have to FOIL at this point. That was done already earlier.

Answer 33

Okay.

Answer 34

After factoring, \(x^2 - 4x - 5\) you set each factor equal to zero, so

\(x - 5 = 0 \)
\(x + 1 = 0\)

Answer 35

So the Coordinates are (0, 0)?

Answer 36

No, the zeroes are an application of the zero product property: \(ab = 0\). You have to solve for x for each factor.

Answer 37

Oh.

Answer 38

I'm still, So confused. Can we Start over? Step by step? because I need to Show my work. I've been on this problem for a while now.

Answer 39

okay, re-post it as a new problem. I will walk you through it this time.

Answer 40

Okay..