Question

There must be a variable term in both the numerator and denominator.
(5x)
(2x)
(x + 4)
(x – 5)
(2x + 1)
(3x + 5)

Answer 1

https://vcsohio.brainhoney.com/Resource/22138800,AB4,0,B,7/Assets/ch5quiz.PNG

Answer 2

@iGreen @ganeshie8

Answer 3

Turn one. Flip your coin and perform the appropriate operation. Explain to the game master how to add your rational expression to the one on the correct space. Use complete sentences.

Answer 4

I got tails

Answer 5

Right..so that means you do:

\(\dfrac{x-1}{3x}\)

Answer 6

That's it?

Answer 7

Actually it wants you do add both of them..so I guess it's:

\((x - 1) + (3x)\)

To simplify, add the like terms..

Answer 8

4x-1?

Answer 9

Yep.

Answer 10

That's the answer? Are you serious?!

Answer 11

Yes.

Answer 12

That's the answer for Turn 1, if you get Tails..

Answer 13

It says I have to explain to the game master how to add your rational expression to the one on the correct space.

Answer 14

Huh..?

Answer 15

Ohhhh..you have to make a rational expression..then you add it with \(\dfrac{x-1}{3x}\).

Answer 16

But what do I do with the given examples (5x)
(2x)
(x + 4)
(x – 5)
(2x + 1)
(3x + 5)

Answer 17

It says there has to be a numerator in the top and bottom..so make one out of the one's they gave you.

Answer 18

Just have '5x' or '2x' be the denominator, and the others the numerator.

Answer 19

so it would be \[\frac{ 4x-1 }{ 2x }\]

Answer 20

No..you have to pick a numerator out of:

(x + 4)
(x - 5)
(2x + 1)
(3x + 5)

Answer 21

Which one do you want to pick?

Answer 22

the third, but why can't I have 2x or 5x be the denominator?

Answer 23

You can..but you just can't have \(4x - 1\) as the numerator..

Answer 24

So therefore our expression is:

\(\dfrac{2x + 1}{2x}\)

Now since you got Tails, we can add this to \(\dfrac{x-1}{3x}\)

So can you add this?

\(\dfrac{2x + 1}{2x} + \dfrac{x - 1}{3x}\)

Answer 25

\[\frac{ 8x+1 }{ 6x }\]?

Answer 26

You got it.

Answer 27

Now flip the coin again.

Answer 28

Tails again

Answer 29

Turn two. Flip your coin and perform the appropriate operation. Discuss and identify any possible restrictions that exist with (or in) the resulting rational expression.

Answer 30

@iGreen

Answer 31

Okay, now we have to subtract.

\(\dfrac{2x + 1}{2x} - \dfrac{x}{x-4}\)

Answer 32

Can you subtract that?

Answer 33

\[-\frac{ 7x+4 }{ 2x(x-4) }?\]

Answer 34

@iGreen

Answer 35

Yep.

Answer 36

Now flip the coin again.

Answer 37

I need to find any restrictions that exist with, or in, the resulting expression before I can move on.

Answer 38

What do you mean by restrictions?

Answer 39

Anything that could prevent the following outcome from presenting itself.

Answer 40

Hmm..I don't see any of those :l

Answer 41

So I'll just put 'No identified restrictions'

Answer 42

Heads: Turn three. Flip your coin and perform the appropriate operation. Explain to the game master how to multiply your rational expression to the one on the correct space. Use complete sentences.

Answer 43

Okay, so now we multiply:

\(\dfrac{2x + 1}{2x} \times \dfrac{x-4}{x+1}\)

What do you get?

Answer 44

\[\frac{ (2x+1)(x-4) }{ 2x(x+1) }\]

Answer 45

@iGreen

Answer 46

Yes..and flip the coin again..

Answer 47

Tails: Turn four. Flip your coin and perform the appropriate operation. Discuss why the degree of the resulting denominator did not change from your expression’s degree.

Answer 48

Okay, so we divide:

\(\dfrac{2x + 1}{2x} \div \dfrac{6}{x-1}\)

Answer 49

\[\frac{ (2x+1)(x-1) }{ 12x }\]

Answer 50

Yep..now we move onto the final step..

Answer 51

I guess we add:

\(\dfrac{2x+1}{2x} + \dfrac{x}{x+1} + \dfrac{x}{x-1}\)

Answer 52

Before we can move on, I need to discover why the degree of the resulting denominator did not change from my expression's degree.

Answer 53

@iGreen

Answer 54

Because when you divided it you got the same degree..

Answer 55

(Which is a degree of 1)

Answer 56

You put dots at the end as if it is a stupid question lol

Answer 57

\[\frac{ 6x^3+x^2-2x-1 }{ 2x(x+1)(x-1) }\]

Answer 58

for the last one ^^^

Answer 59

@iGreen

Answer 60

Yep, you got it.!

Answer 61

Thanks so much!

Answer 62

No problem, you did pretty much all the work :P

Answer 63

You were very patient with my incompetence though.