ElisaNeedsHelp
  • ElisaNeedsHelp
Multiplying rational expressions involving multivariate monomials, I'll give you a medal Please
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ElisaNeedsHelp
  • ElisaNeedsHelp
\[ \frac{ 6x}{ 2b}\times \frac{ 8bx }{ 3x}\]
ElisaNeedsHelp
  • ElisaNeedsHelp
I understand the basic concept of what to do but I keep getting it wrong since every problem has its little difference
Jhannybean
  • Jhannybean
How did you initially go about solving it?

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ElisaNeedsHelp
  • ElisaNeedsHelp
well I multiply and I get 48/6 and i cancel out the variables this time?
Jhannybean
  • Jhannybean
Well, if you treat \(x\), and \(b\), both as variables, would you not be able to cancel them out via cross multiplication?
Jhannybean
  • Jhannybean
Cancel like terms with like terms
ElisaNeedsHelp
  • ElisaNeedsHelp
Can we try another one ? I would continue with this one but my electronic workbook kicked me off that problem
Jhannybean
  • 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
  • Jhannybean
Sure.
ElisaNeedsHelp
  • ElisaNeedsHelp
Sorry and thanks.\[\frac{ 3a }{ 2x}\times \frac{ 4x ^{5} }{ 9ax}\]
ElisaNeedsHelp
  • ElisaNeedsHelp
That is a 4x^5
Jhannybean
  • 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
  • Jhannybean
so comparing all the like colored terms together, see how we can cross cancel the blue a terms?
ElisaNeedsHelp
  • ElisaNeedsHelp
is it okay if I multiply first and then cross them out? or I shouldnt do that in the future?
ElisaNeedsHelp
  • ElisaNeedsHelp
so far I have \[\frac{ 12 }{ 18x ^{3} }\]
Jhannybean
  • 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
  • Jhannybean
if you're multiplying them all out, what happened to your \(a\) terms?
Jhannybean
  • Jhannybean
Oh I see, so you've partially reduced your function.
ElisaNeedsHelp
  • ElisaNeedsHelp
Jhannybean
  • 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
  • ElisaNeedsHelp
Wow and could the same thing be done for \[\frac{ 15y }{ 2b ^{2} }\times \frac{ b ^{3} y}{ 9y ^{4} }\]
ElisaNeedsHelp
  • ElisaNeedsHelp
well not exactly the same which is what i hate
Jhannybean
  • Jhannybean
It will work the same way :) yes!
ElisaNeedsHelp
  • ElisaNeedsHelp
let me try now
ElisaNeedsHelp
  • ElisaNeedsHelp
do I simplify before I multiply?
Jhannybean
  • Jhannybean
Yes :)
ElisaNeedsHelp
  • ElisaNeedsHelp
Do I only simplify the 15 and the 9 into 3s? or am i completely wrong?
ElisaNeedsHelp
  • ElisaNeedsHelp
no the 15 into a 5 and the 9 into a 3
ElisaNeedsHelp
  • ElisaNeedsHelp
I need guidance I'm confused especially with the variables
Jhannybean
  • Jhannybean
yes you're correct, simplify them both by the LCM
Jhannybean
  • Jhannybean
\[\huge \frac{ \color{red}{5}\cancel{15}y }{ 2b ^{2} }\times \frac{ b ^{3} y}{ \color{red}{3}\cancel{9}y ^{4} }\]
ElisaNeedsHelp
  • ElisaNeedsHelp
I also have confusion on where to place each variable on the final answer. I have plain b and y^6
Jhannybean
  • 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
  • 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
  • ElisaNeedsHelp
Lol but the variables are everywhere. Maybe number 2 is the best for me
Jhannybean
  • Jhannybean
Alright
Jhannybean
  • 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
  • ElisaNeedsHelp
Yeah I see Thats my safe zone
Jhannybean
  • Jhannybean
:) alright
Jhannybean
  • 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
  • ElisaNeedsHelp
yes :)
Jhannybean
  • 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
  • ElisaNeedsHelp
b and y^-2
Jhannybean
  • 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
  • ElisaNeedsHelp
Yeap following
Jhannybean
  • Jhannybean
therefore we can rewrite our function as: \[\huge \frac{5}{6} \cdot b \cdot \frac{1}{y^2} = \frac{5b}{6y^2}\]
ElisaNeedsHelp
  • ElisaNeedsHelp
I got your message but OS wont let me reply but its alright :P lol
Jhannybean
  • Jhannybean
Do you understand this better, now? :D
ElisaNeedsHelp
  • ElisaNeedsHelp
Yes I'll try doing the other one completely by myself!
Jhannybean
  • Jhannybean
Yay!
ElisaNeedsHelp
  • ElisaNeedsHelp
Am I right so far? on #3
ElisaNeedsHelp
  • ElisaNeedsHelp
Lol i tried to do it all by myself
ElisaNeedsHelp
  • ElisaNeedsHelp
is the final answer\[\frac{ 12b ^{2} }{ y ^{4} }\]
ElisaNeedsHelp
  • ElisaNeedsHelp
No its not but i tried
Jhannybean
  • Jhannybean
number 2 seems wrong.... we worked this out just now though ^^
ElisaNeedsHelp
  • ElisaNeedsHelp
Lol yeah dont worry I havent erased but I will go through our thread and rewrite it
Jhannybean
  • Jhannybean
Oh okay
Jhannybean
  • 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
  • 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
  • ElisaNeedsHelp
Thanks for the help Jhannybean ! :)
Jhannybean
  • Jhannybean
no problem:)

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