Question

x = y^2 - 1
2x - y = 4

Which of the following equations could be the result of using substitution to solve the system?
2y^2 - y - 5 = 0
2y^2 - y - 6 = 0
2y^2 - 2y - 5 = 0

Answer 1

\[x = y^{2}-1 \] see this right?

Answer 2

Yes

Answer 3

think of this as saying x = y^2-1

so wherever you see x you plug in y^2-1

where can you do this?

Answer 4

For the second equation in the system it would be 2(y^2-1) = y = 4

Answer 5

almost

Answer 6

\[2(y^{2}-1)-y = 4\]

Answer 7

Ah, I see the mistype.

Answer 8

know how to continue?

Answer 9

Not fully.

Answer 10

see that 2 out side ?

Answer 11

\[2(~~~~) \]

Answer 12

Yes.

Answer 13

\[2y^{2}-2-y = 4 \]

Answer 14

follow?

Answer 15

@OracleThinkTank

Answer 16

and also how would we continue based off of this

Answer 17

Okay.

Answer 18

yeah so how do you think we would continue?

Answer 19

I'm not too sure. I'm not the best with Algebra.

Answer 20

so the next step is we move all the terms to the same side.

Answer 21

so we have a 4 on one side right? and also, this is a quadratic equation

Answer 22

Okay.

Answer 23

\[2y^{2}-y-2 = 4 \]

Answer 24

now we subtract 4 from both sides.

\[2y^{2}-y-2-4 = 0\]

how would we simplify this?

Answer 25

3y^2 - 6 = 0?

Answer 26

\[2y^{2}-y-6 = 0 \]

remember you can only combine like terms.

Answer 27

Oh, Thanks.