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ElisaNeedsHelp

  • one year ago

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

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  1. ElisaNeedsHelp
    • one year ago
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    \[ \frac{ 6x}{ 2b}\times \frac{ 8bx }{ 3x}\]

  2. ElisaNeedsHelp
    • one year ago
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    I understand the basic concept of what to do but I keep getting it wrong since every problem has its little difference

  3. Jhannybean
    • one year ago
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    How did you initially go about solving it?

  4. ElisaNeedsHelp
    • one year ago
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    well I multiply and I get 48/6 and i cancel out the variables this time?

  5. Jhannybean
    • one year ago
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    Well, if you treat \(x\), and \(b\), both as variables, would you not be able to cancel them out via cross multiplication?

  6. Jhannybean
    • one year ago
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    Cancel like terms with like terms

  7. ElisaNeedsHelp
    • one year ago
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    Can we try another one ? I would continue with this one but my electronic workbook kicked me off that problem

  8. Jhannybean
    • one year ago
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    \[\huge \frac{\color{orange}{6}\color{red}{x}}{\color{orange}{2}\color{blue}{b}} \cdot \frac{\color{orange}{8}\color{blue}{b}\color{red}{x}}{\color{orange}{3}\color{red}{x}}\]

  9. Jhannybean
    • one year ago
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    Sure.

  10. ElisaNeedsHelp
    • one year ago
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    Sorry and thanks.\[\frac{ 3a }{ 2x}\times \frac{ 4x ^{5} }{ 9ax}\]

  11. ElisaNeedsHelp
    • one year ago
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    That is a 4x^5

  12. Jhannybean
    • one year ago
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    \[\huge \frac{\color{orange}{3}\color{blue}{a}}{\color{orange}{2}\color{red}{x}} \cdot \frac{\color{orange}{4}\color{red}{x^5}}{\color{orange}{9}\color{blue}{a}\color{red}{x}}\]

  13. Jhannybean
    • one year ago
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    so comparing all the like colored terms together, see how we can cross cancel the blue a terms?

  14. ElisaNeedsHelp
    • one year ago
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    is it okay if I multiply first and then cross them out? or I shouldnt do that in the future?

  15. ElisaNeedsHelp
    • one year ago
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    so far I have \[\frac{ 12 }{ 18x ^{3} }\]

  16. Jhannybean
    • one year ago
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    it's easier to cross cancel terms that can be simplified first because sometimes you'll get gigantic numbers that will overwhelm you.

  17. Jhannybean
    • one year ago
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    if you're multiplying them all out, what happened to your \(a\) terms?

  18. Jhannybean
    • one year ago
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    Oh I see, so you've partially reduced your function.

  19. ElisaNeedsHelp
    • one year ago
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  20. Jhannybean
    • one year ago
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    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 :\[\huge \frac{2}{3}x^{5-2} = \frac{2}{3} x^3\]

  21. ElisaNeedsHelp
    • one year ago
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    Wow and could the same thing be done for \[\frac{ 15y }{ 2b ^{2} }\times \frac{ b ^{3} y}{ 9y ^{4} }\]

  22. ElisaNeedsHelp
    • one year ago
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    well not exactly the same which is what i hate

  23. Jhannybean
    • one year ago
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    It will work the same way :) yes!

  24. ElisaNeedsHelp
    • one year ago
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    let me try now

  25. ElisaNeedsHelp
    • one year ago
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    do I simplify before I multiply?

  26. Jhannybean
    • one year ago
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    Yes :)

  27. ElisaNeedsHelp
    • one year ago
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    Do I only simplify the 15 and the 9 into 3s? or am i completely wrong?

  28. ElisaNeedsHelp
    • one year ago
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    no the 15 into a 5 and the 9 into a 3

  29. ElisaNeedsHelp
    • one year ago
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    I need guidance I'm confused especially with the variables

  30. Jhannybean
    • one year ago
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    yes you're correct, simplify them both by the LCM

  31. Jhannybean
    • one year ago
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    \[\huge \frac{ \color{red}{5}\cancel{15}y }{ 2b ^{2} }\times \frac{ b ^{3} y}{ \color{red}{3}\cancel{9}y ^{4} }\]

  32. ElisaNeedsHelp
    • one year ago
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    I also have confusion on where to place each variable on the final answer. I have plain b and y^6

  33. Jhannybean
    • one year ago
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    You keep the variables where they are unless they can be reduced and canceled out, similar to our last problem if you scrolled up ^^

  34. Jhannybean
    • one year ago
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    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?

  35. ElisaNeedsHelp
    • one year ago
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    Lol but the variables are everywhere. Maybe number 2 is the best for me

  36. Jhannybean
    • one year ago
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    Alright

  37. Jhannybean
    • one year ago
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    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? ^^

  38. ElisaNeedsHelp
    • one year ago
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    Yeah I see Thats my safe zone

  39. Jhannybean
    • one year ago
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    :) alright

  40. Jhannybean
    • one year ago
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    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

  41. ElisaNeedsHelp
    • one year ago
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    yes :)

  42. Jhannybean
    • one year ago
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    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

  43. ElisaNeedsHelp
    • one year ago
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    b and y^-2

  44. Jhannybean
    • one year ago
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    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?

  45. ElisaNeedsHelp
    • one year ago
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    Yeap following

  46. Jhannybean
    • one year ago
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    therefore we can rewrite our function as: \[\huge \frac{5}{6} \cdot b \cdot \frac{1}{y^2} = \frac{5b}{6y^2}\]

  47. ElisaNeedsHelp
    • one year ago
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    I got your message but OS wont let me reply but its alright :P lol

  48. Jhannybean
    • one year ago
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    Do you understand this better, now? :D

  49. ElisaNeedsHelp
    • one year ago
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    Yes I'll try doing the other one completely by myself!

  50. Jhannybean
    • one year ago
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    Yay!

  51. ElisaNeedsHelp
    • one year ago
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    Am I right so far? on #3

  52. ElisaNeedsHelp
    • one year ago
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    Lol i tried to do it all by myself

  53. ElisaNeedsHelp
    • one year ago
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    is the final answer\[\frac{ 12b ^{2} }{ y ^{4} }\]

  54. ElisaNeedsHelp
    • one year ago
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    No its not but i tried

  55. Jhannybean
    • one year ago
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    number 2 seems wrong.... we worked this out just now though ^^

  56. ElisaNeedsHelp
    • one year ago
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    Lol yeah dont worry I havent erased but I will go through our thread and rewrite it

  57. Jhannybean
    • one year ago
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    Oh okay

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

  59. Jhannybean
    • one year ago
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    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!!

  60. ElisaNeedsHelp
    • one year ago
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    Thanks for the help Jhannybean ! :)

  61. Jhannybean
    • one year ago
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    no problem:)

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