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ElisaNeedsHelp
 one year ago
Multiplying rational expressions involving multivariate monomials, I'll give you a medal Please
ElisaNeedsHelp
 one year ago
Multiplying rational expressions involving multivariate monomials, I'll give you a medal Please

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ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0\[ \frac{ 6x}{ 2b}\times \frac{ 8bx }{ 3x}\]

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0I understand the basic concept of what to do but I keep getting it wrong since every problem has its little difference

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How did you initially go about solving it?

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0well I multiply and I get 48/6 and i cancel out the variables this time?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, if you treat \(x\), and \(b\), both as variables, would you not be able to cancel them out via cross multiplication?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Cancel like terms with like terms

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0Can we try another one ? I would continue with this one but my electronic workbook kicked me off that problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge \frac{\color{orange}{6}\color{red}{x}}{\color{orange}{2}\color{blue}{b}} \cdot \frac{\color{orange}{8}\color{blue}{b}\color{red}{x}}{\color{orange}{3}\color{red}{x}}\]

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0Sorry and thanks.\[\frac{ 3a }{ 2x}\times \frac{ 4x ^{5} }{ 9ax}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge \frac{\color{orange}{3}\color{blue}{a}}{\color{orange}{2}\color{red}{x}} \cdot \frac{\color{orange}{4}\color{red}{x^5}}{\color{orange}{9}\color{blue}{a}\color{red}{x}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so comparing all the like colored terms together, see how we can cross cancel the blue a terms?

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0is it okay if I multiply first and then cross them out? or I shouldnt do that in the future?

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0so far I have \[\frac{ 12 }{ 18x ^{3} }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's easier to cross cancel terms that can be simplified first because sometimes you'll get gigantic numbers that will overwhelm you.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you're multiplying them all out, what happened to your \(a\) terms?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I see, so you've partially reduced your function.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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 :\[\huge \frac{2}{3}x^{52} = \frac{2}{3} x^3\]

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0Wow and could the same thing be done for \[\frac{ 15y }{ 2b ^{2} }\times \frac{ b ^{3} y}{ 9y ^{4} }\]

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0well not exactly the same which is what i hate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It will work the same way :) yes!

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0do I simplify before I multiply?

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0Do I only simplify the 15 and the 9 into 3s? or am i completely wrong?

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0no the 15 into a 5 and the 9 into a 3

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0I need guidance I'm confused especially with the variables

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes you're correct, simplify them both by the LCM

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge \frac{ \color{red}{5}\cancel{15}y }{ 2b ^{2} }\times \frac{ b ^{3} y}{ \color{red}{3}\cancel{9}y ^{4} }\]

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0I also have confusion on where to place each variable on the final answer. I have plain b and y^6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You keep the variables where they are unless they can be reduced and canceled out, similar to our last problem if you scrolled up ^^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0Lol but the variables are everywhere. Maybe number 2 is the best for me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now 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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I see Thats my safe zone

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now we apply the subtraction rule for exponents, and we leave whatever fractions that cannot be simplified any further alone. \[\huge \frac{5}{6} \cdot b^{32} \cdot y^{24}\] What does this give you when simplified, can you tell me? :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Good. 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0therefore we can rewrite our function as: \[\huge \frac{5}{6} \cdot b \cdot \frac{1}{y^2} = \frac{5b}{6y^2}\]

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0I got your message but OS wont let me reply but its alright :P lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you understand this better, now? :D

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0Yes I'll try doing the other one completely by myself!

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0Am I right so far? on #3

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0Lol i tried to do it all by myself

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0is the final answer\[\frac{ 12b ^{2} }{ y ^{4} }\]

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0No its not but i tried

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0number 2 seems wrong.... we worked this out just now though ^^

ElisaNeedsHelp
 one year ago
Best ResponseYou've already chosen the best response.0Lol yeah dont worry I havent erased but I will go through our thread and rewrite it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align} \huge \frac{8b^3y^4}{3y} \cdot \frac{9y}{2b} &= \huge \frac{72}{6}\cdot \frac{b^3}{b}\cdot \frac{y^5}{y} \\ &= \huge \frac{72}{6} \cdot b^{31} \cdot y^{51} \\&= \huge ....?\end{align}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Try 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
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the help Jhannybean ! :)
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