Question

Please help will fan and medal :(
What is the solution set of 4x = y and 2x2 − y = 0?

{(0.5, -0.5), (2, -8)}

{(0, 0), (2, 8)}

{(1.5, 2), (8, 2)}

{(-1.5, -2), (2, 8)}

Answer 1

@dan815 @Owlcoffee

Answer 2

Hello! Do you know how to use the substitution method?

Answer 3

Like plugging them in?

Answer 4

Yes! We know what y is (4x), so let's plug that in to the second equation :)

Answer 5

Erm, how so ._.,

Answer 6

Is that supposed to be 2x^2?

Answer 7

Okay, so you see that y = 4x right? the first equation

Answer 8

Yes 2x^2 and I see it

Answer 9

Okay, so instead of y in the second equation, we're going to put in 4x in place of that y

Answer 10

Ahh okay. so then from there where would we go?

Answer 11

Okay, so now we have, \(2x^2 -4y=0\), right?
Let me make sure I'm right on the next step, one minute please :)

Answer 12

*4x

Answer 13

Okay, so do you know what the quadratic formula is?

Answer 14

ax+bx=c^2?

Answer 15

Actually, we have a^2 + bx + c = 0, c would be 0, we need to put those values into the quadratic formula (this may look scary but it's really not), \(x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Answer 16

and because c is 0, that makes it even easier :)

Answer 17

Okay one second while I try ._.

Answer 18

Uhm, I'm a bit stuck ._. @sleepyjess

Answer 19

Okay, well we know a is 2, and b is -4, c is 0 because there is no c in our equation :)
x = \(\dfrac{4\pm\sqrt{-4^2-4(2)(0)}}{2(2)}\)
and now, we just work it all out

Answer 20

Okay, now I'm questioning myself...

Answer 21

4x = y and 2x2 − y = 0

let me just do it step-by-step
1. sub y= 4x into the 2nd equation: \(\sf 2x^2-4x=0\)
2. Solve for 'x' by quadratic formula (a=4, b=-4, c=0) or by factoring (which is much easier)

I think jess can continue it from here now ^_^

Answer 22

(Sprry my computer froze)

Answer 23

\(\sf x= \dfrac{4\pm\sqrt{(-4)^2-4(2)(0)}}{2(2)}\)

Answer 24

\(\sf x=\dfrac{4\pm\sqrt{16}}{4}\)

Answer 25

now what is square root of 16? @WhatEven

Answer 26

4!

Answer 27

but none of the options are 4 @Jadzia

Answer 28

right..we are not done yet. we are only working on the radical part

so now it will be like this:
\(\sf x=\dfrac{4\pm 4}{4}\)

which we will split into two parts:

\(\sf x=\dfrac{4+ 4}{4}\) and \(\sf x=\dfrac{4- 4}{4}\)

now what values are you going to get for x if you solve those two?

Answer 29

2 and -2?

Answer 30

2 is right, -2 is not...

what is \(\sf x=\dfrac{4-4}{4}\)?

Answer 31

@whateven still there?

Answer 32

after you find the value of x on the second one.. the next step is to plug those x values on to the equation to solve for their corresponding y-components.

Answer 33

Computer froze again ;/

Answer 34

try restarting your computer

Answer 35

I will if it does it again but it glitches and takes an hour to do so sometimes so I'd rather not risk that yet. I got 0

Answer 36

@Jadzia still there?

Answer 37

yes that's right.
So from your choices, which them gives you 2 and 0 as x-coordinates?

Answer 38

**of