Question

Need some algebra 2 help, please! Would be highly appreciated

Answer 1

1.Create a rational expression to be your game piece. You may choose from the list of factors below or make your own. There must be a variable term in both the numerator and denominator.
•(5x)
•(2x)
•(x + 4)
•(x – 5)
•(2x + 1)
•(3x + 1)
I"m not sure as to what rational I should make. I was thinking of using the (2x)

Answer 2

@agent0smith @jim_thompson5910 Can one of you help please

Answer 3

Simply pick two expressions and divide them. That's how you'll create a rational expression.

Answer 4

Just choose a couple... I pick (2x)/(3x + 1)

Answer 5

Well, the trick is that it's like a "board game" so I need to flip a coin, and whatevr side it lands on I have to solve it using the expression i guess?

Answer 6

seems like you start with 0

then if you flip heads on the first coin flip, then you add on \(\Large \frac{x+1}{2x}\) to 0

or if you flip tails on the first coin flip, then you add on \(\Large \frac{x-1}{3x}\) to 0

Answer 7

I think we have to create one to add it to. Like we start out with our created "game piece" and if i flip tails then I have to solve my rational experession with the other.

Answer 8

oh i see, that works too

Answer 9

So we could use Agent's expression? Like I flipped heads so it would be\[ \frac{ 2x }{ 3x+1 } + \frac{ x+1 }{ 2x }\]

Answer 10

yes, then you would add and simplify

Answer 11

Could you explain how to for me? So I can do it for the next 4 questions?

Answer 12

in general, how do you add fractions

let's say you wanted to add \(\Large \frac{1}{2}+\frac{1}{3}\)

how would you do it?

Answer 13

Find a common denominator?

Answer 14

which is what?

Answer 15

Six. Times the 2 by 3 and times the 3 by 2, right?

Answer 16

good

Answer 17

in the first fraction, you multiplied the denominator by 3

so you have to do the same to the numerator to balance it out

so

\[\Large \frac{1}{2} = \frac{1*3}{2*3} = \frac{3}{6}\]

see how I'm doing this?

Answer 18

Yes, I'm following along :)

Answer 19

Similarly,

\[\Large \frac{1}{3} = \frac{1*2}{3*2} = \frac{2}{6}\]

Answer 20

which means

\[\Large \frac{1}{2}+\frac{1}{3}\]

is the same as

\[\Large \frac{3}{6}+\frac{2}{6}\]

Answer 21

now that the denominators are the same, you can add the numerators and place them over the common denominator

Answer 22

so you apply these basic steps to add rational expressions

Answer 23

So since my denominators have a 3 and a 2, would it be 6 again for this as well? It's confusing when it haves the 3x+1

Answer 24

in that last example, the denominators were 2 and 3

the LCD is 6 since 2*3 = 6 (and the GCF of 2 and 3 is 1)

Answer 25

so if there is no common factor between the denominators (other than 1), you just multiply the denominators to get the LCD

Answer 26

So I would muiltiply 3x+1 + 2y? sorry if i'm frustrating you i'm just really confused

Answer 27

the denominators in this case are 3x+1 and 2x

multiply them to get 2x(3x+1)

you can leave it factored or you can distribute

Answer 28

so the LCD is 2x(3x+1)

Answer 29

the goal is to get every denominator equal to the LCD

Answer 30

So it would now be
\[\frac{ 2x }{2x(3x+1)} + \frac{ x+1 }{ 2x(3x+1) }\]

Answer 31

you have the denominators correct, but the numerators are incorrect

Answer 32

in the first fraction, you need to multiply top and bottom by 2x to get


\[\Large \frac{ 2x }{ 3x+1 }\]


\[\Large \frac{ 2x*2x }{ 2x(3x+1) }\]


\[\Large \frac{ 2*2*x*x }{ 2x(3x+1) }\]


\[\Large \frac{ 4x^2 }{ 2x(3x+1) }\]


You have to multiply top and bottom of the fraction to balance it out

Answer 33

So since I rolled heads and we found the LCD would this be the answer for this question. Explain to the game master how to add your rational expression to the one on the correct space. Or do we still have to add together now?

Answer 34

no you need to add the rational expressions

Answer 35

Okay, so how could we do that?

Answer 36

you get each fraction to get the LCD using the methods shown above

Answer 37

once the fractions have the same denominator, you can add the numerators and place them over the LCD

Answer 38

So 4x^2 + 4x2 would be 8x^2 or 32x?

Answer 39

for the second fraction, you have to multiply top and bottom by 3x+1

Answer 40

What do you mean

Answer 41

\[\Large \frac{ x+1 }{ 2x }\]


\[\Large \frac{ (3x+1)(x+1) }{ (3x+1)2x }\]


\[\Large \frac{ 3x(x+1)+1(x+1) }{ 2x(3x+1) }\]


I'll let you finish simplifying

Answer 42

I'm stumped man, I'm trying to figure it out I promise it's just super confusing. So would I do \[\frac{ 3x(x+1 + 1 (x-1) }{ 2x (3x +1) }\]
And then \[4x + 1 (x+1) \] For the numerator? Sorry if i'm totally failing

Answer 43

no

Answer 44

3x(x+1) + 1(x+1)

3x*x + 3x*1 + 1*x + 1*1

3x^2 + 3x + 1x + 1

3x^2 + 4x + 1

Answer 45

so,

\[\Large \frac{ 2x }{ 3x+1 } + \frac{ x+1 }{ 2x }\]

is the same as

\[\Large \frac{ 4x^2 }{ 2x(3x+1) } + \frac{ 3x^2+4x+1 }{ 2x(3x+1) }\]

Answer 46

So now how do we solve?

Answer 47

the denominators are now equal, so you can add the numerators and place that result over the LCD

Answer 48

Are we doing a combining like terms kinda thing?

Answer 49

yes

Answer 50

so \[\frac{ 3^{x2 } + 4x +1}{2x(3x+1 }\]

Answer 51

no

Answer 52

:/ sorry

Answer 53

3x^2 + 4x^2 is ????

Answer 54

7x^2

Answer 55

\[\Large \frac{ 2x }{ 3x+1 } + \frac{ x+1 }{ 2x }\]


\[\Large \frac{ 4x^2 }{ 2x(3x+1) } + \frac{ 3x^2+4x+1 }{ 2x(3x+1) }\]


\[\Large \frac{ 4x^2 + 3x^2+4x+1 }{ 2x(3x+1) }\]


\[\Large \frac{ 7x^2+4x+1 }{ 2x(3x+1) }\]

Answer 56

Thank you! Sorry it took me a long time

Answer 57

that's ok, and yw