Question

Multiplying rational expressions involving multivariate monomials, I'll give you a medal Please

Answer 1

\[ \frac{ 6x}{ 2b}\times \frac{ 8bx }{ 3x}\]

Answer 2

I understand the basic concept of what to do but I keep getting it wrong since every problem has its little difference

Answer 3

How did you initially go about solving it?

Answer 4

well I multiply and I get 48/6 and i cancel out the variables this time?

Answer 5

Well, if you treat \(x\), and \(b\), both as variables, would you not be able to cancel them out via cross multiplication?

Answer 6

Cancel like terms with like terms

Answer 7

Can we try another one ? I would continue with this one but my electronic workbook kicked me off that problem

Answer 8

\[\huge \frac{\color{orange}{6}\color{red}{x}}{\color{orange}{2}\color{blue}{b}} \cdot \frac{\color{orange}{8}\color{blue}{b}\color{red}{x}}{\color{orange}{3}\color{red}{x}}\]

Answer 9

Sure.

Answer 10

Sorry and thanks.\[\frac{ 3a }{ 2x}\times \frac{ 4x ^{5} }{ 9ax}\]

Answer 11

That is a 4x^5

Answer 12

\[\huge \frac{\color{orange}{3}\color{blue}{a}}{\color{orange}{2}\color{red}{x}} \cdot \frac{\color{orange}{4}\color{red}{x^5}}{\color{orange}{9}\color{blue}{a}\color{red}{x}}\]

Answer 13

so comparing all the like colored terms together, see how we can cross cancel the blue a terms?

Answer 14

is it okay if I multiply first and then cross them out? or I shouldnt do that in the future?

Answer 15

so far I have \[\frac{ 12 }{ 18x ^{3} }\]

Answer 16

it's easier to cross cancel terms that can be simplified first because sometimes you'll get gigantic numbers that will overwhelm you.

Answer 17

if you're multiplying them all out, what happened to your \(a\) terms?

Answer 18

Oh I see, so you've partially reduced your function.

Answer 19

So how I would solve it is just cross cancel everything I can simplify: \[\huge \frac{\cancel{\color{orange}{3}}\color{blue}{a}}{\cancel{\color{orange}{2}}\color{red}{x}} \cdot \frac{2\cancel{\color{orange}{4}}\color{red}{x^5}}{3\cancel{\color{orange}{9}}\color{blue}{a}\color{red}{x}}\] this leaves me with: \[\huge \frac{2\color{blue}{a}\color{red}{x^5}}{3\color{blue}{a}\color{red}{x^2}}\]Now i can cancel out by a's and i'll be left with : \[\huge \frac{2\color{red}{x^5}}{3\color{red}{x^2}}\]using the power subtraction rule for exponents, I will end up with my final answer being :\[\huge \frac{2}{3}x^{5-2} = \frac{2}{3} x^3\]

Answer 20

Wow and could the same thing be done for \[\frac{ 15y }{ 2b ^{2} }\times \frac{ b ^{3} y}{ 9y ^{4} }\]

Answer 21

well not exactly the same which is what i hate

Answer 22

It will work the same way :) yes!

Answer 23

let me try now

Answer 24

do I simplify before I multiply?

Answer 25

Yes :)

Answer 26

Do I only simplify the 15 and the 9 into 3s? or am i completely wrong?

Answer 27

no the 15 into a 5 and the 9 into a 3

Answer 28

I need guidance I'm confused especially with the variables

Answer 29

yes you're correct, simplify them both by the LCM

Answer 30

\[\huge \frac{ \color{red}{5}\cancel{15}y }{ 2b ^{2} }\times \frac{ b ^{3} y}{ \color{red}{3}\cancel{9}y ^{4} }\]

Answer 31

I also have confusion on where to place each variable on the final answer. I have plain b and y^6

Answer 32

You keep the variables where they are unless they can be reduced and canceled out, similar to our last problem if you scrolled up ^^

Answer 33

So with this problem we can do one of two things:

1. cross multiply and cancel out the \(y\)'s that can be cancelled
2. multiply across the numerator and denominator and worry about cancelling afterward

Which would you prefer?

Answer 34

Lol but the variables are everywhere. Maybe number 2 is the best for me

Answer 35

Alright

Answer 36

Now that we've cross multiplied and reduced the numbers, let's multiply it out :) \[\huge \frac{5y^2b^3}{6b^2y^4}\] I didnt see the other b there at first! :o

Do you see how I got this fraction? ^^

Answer 37

Yeah I see Thats my safe zone

Answer 38

:) alright

Answer 39

Now let's separate each like term into a fraction, this will help us use the subtraction rule for exponents \[\huge \frac{5}{6} \cdot \frac{b^3}{b^2} \cdot \frac{y^2}{y^4}\] Are you comfortable seeign it like this? :o

Answer 40

yes :)

Answer 41

Now we apply the subtraction rule for exponents, and we leave whatever fractions that cannot be simplified any further alone. \[\huge \frac{5}{6} \cdot b^{3-2} \cdot y^{2-4}\] What does this give you when simplified, can you tell me? :P

Answer 42

b and y^-2

Answer 43

Good. We want t write al our powers as positive powers, therefore we have to change \(y^{-2}\) to \(\dfrac{1}{y^2}\) to keep it positive. :) Following?

Answer 44

Yeap following

Answer 45

therefore we can rewrite our function as: \[\huge \frac{5}{6} \cdot b \cdot \frac{1}{y^2} = \frac{5b}{6y^2}\]

Answer 46

I got your message but OS wont let me reply but its alright :P lol

Answer 47

Do you understand this better, now? :D

Answer 48

Yes I'll try doing the other one completely by myself!

Answer 49

Yay!

Answer 50

Am I right so far? on #3

Answer 51

Lol i tried to do it all by myself

Answer 52

is the final answer\[\frac{ 12b ^{2} }{ y ^{4} }\]

Answer 53

No its not but i tried

Answer 54

number 2 seems wrong.... we worked this out just now though ^^

Answer 55

Lol yeah dont worry I havent erased but I will go through our thread and rewrite it

Answer 56

Oh okay

Answer 57

\[\begin{align} \huge \frac{8b^3y^4}{3y} \cdot \frac{9y}{2b} &= \huge \frac{72}{6}\cdot \frac{b^3}{b}\cdot \frac{y^5}{y} \\ &= \huge \frac{72}{6} \cdot b^{3-1} \cdot y^{5-1} \\&= \huge ....?\end{align}\]

Answer 58

Try working it out this way, and see if this method is easier for you. I've got to head off OS right now, so good luck on solving the rest!!

Answer 59

Thanks for the help Jhannybean ! :)

Answer 60

no problem:)