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

- ElisaNeedsHelp

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

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- schrodinger

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- ElisaNeedsHelp

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

- ElisaNeedsHelp

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

- Jhannybean

How did you initially go about solving it?

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## More answers

- ElisaNeedsHelp

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

- Jhannybean

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

- Jhannybean

Cancel like terms with like terms

- ElisaNeedsHelp

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

- Jhannybean

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

- Jhannybean

Sure.

- ElisaNeedsHelp

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

- ElisaNeedsHelp

That is a 4x^5

- Jhannybean

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

- Jhannybean

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

- ElisaNeedsHelp

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

- ElisaNeedsHelp

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

- Jhannybean

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

- Jhannybean

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

- Jhannybean

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

- ElisaNeedsHelp

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- Jhannybean

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 :\[\huge \frac{2}{3}x^{5-2} = \frac{2}{3} x^3\]

- ElisaNeedsHelp

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

- ElisaNeedsHelp

well not exactly the same which is what i hate

- Jhannybean

It will work the same way :) yes!

- ElisaNeedsHelp

let me try now

- ElisaNeedsHelp

do I simplify before I multiply?

- Jhannybean

Yes :)

- ElisaNeedsHelp

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

- ElisaNeedsHelp

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

- ElisaNeedsHelp

I need guidance I'm confused especially with the variables

- Jhannybean

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

- Jhannybean

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

- ElisaNeedsHelp

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

- Jhannybean

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

- Jhannybean

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?

- ElisaNeedsHelp

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

- Jhannybean

Alright

- Jhannybean

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? ^^

- ElisaNeedsHelp

Yeah I see Thats my safe zone

- Jhannybean

:) alright

- Jhannybean

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

- ElisaNeedsHelp

yes :)

- Jhannybean

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

- ElisaNeedsHelp

b and y^-2

- Jhannybean

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?

- ElisaNeedsHelp

Yeap following

- Jhannybean

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

- ElisaNeedsHelp

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

- Jhannybean

Do you understand this better, now? :D

- ElisaNeedsHelp

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

- Jhannybean

Yay!

- ElisaNeedsHelp

Am I right so far? on #3

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- ElisaNeedsHelp

Lol i tried to do it all by myself

- ElisaNeedsHelp

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

- ElisaNeedsHelp

No its not but i tried

- Jhannybean

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

- ElisaNeedsHelp

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

- Jhannybean

Oh okay

- Jhannybean

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

- Jhannybean

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!!

- ElisaNeedsHelp

Thanks for the help Jhannybean ! :)

- Jhannybean

no problem:)

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