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pokemon23

  • 3 years ago

If y = 5x + 2, then find the value of 10xy + 4y in terms of y.

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  1. pokemon23
    • 3 years ago
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    @Hero @Calcmathlete

  2. bahrom7893
    • 3 years ago
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    wow u didn't type @bah here...

  3. bahrom7893
    • 3 years ago
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    I'm not smart enough????

  4. bahrom7893
    • 3 years ago
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    y=5x+2 y-2 = 5x x = (1/5)(y-2)

  5. bahrom7893
    • 3 years ago
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    10xy + 4y = 10 (1/5)(y-2) y + 4y = 2(y - 2)y + 4y = 2y^2 - 4y + 4y = 2y^2 <-Final answer

  6. bahrom7893
    • 3 years ago
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    and it's you're

  7. pokemon23
    • 3 years ago
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    o_o I'm taking the PSAT... I don't know how to do these types of questions......

  8. Calcmathlete
    • 3 years ago
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    To clarify on what @bahrom7893 said, I would've done something a bit simpler? \[\implies10xy + 4y~~~~~~~~~~Original~Expression\]\[\implies 2y(5x + 2)~~~~~~~~~Factor~out~the~GCF\]\[\implies 2y(y) ~~~~~~~~~~~~~~~~~~Substitute\]\[\implies 2y^2~~~~~~~~~~~~~~~~~~~~~Multiplication~of~variables\]

  9. pokemon23
    • 3 years ago
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    ok let's do another problem I need to get better.

  10. Calcmathlete
    • 3 years ago
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    Uh ok then... \[y = (3x + 2)^3\]\[\text{Simplify in terms of y: } (9x + 6)^3\]

  11. Calcmathlete
    • 3 years ago
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    Very similar to the one you were doing on that PSAT question.

  12. pokemon23
    • 3 years ago
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    ok I think I got it.

  13. pokemon23
    • 3 years ago
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    so do I multiply to the 3rd power?

  14. Calcmathlete
    • 3 years ago
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    Yes, the \(^{'3'}\) implies cubing it or multiplying by itself 3 times.

  15. pokemon23
    • 3 years ago
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    9x*9x*9x?

  16. Calcmathlete
    • 3 years ago
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    Not quite...could you show me what you did?

  17. pokemon23
    • 3 years ago
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    well you its |dw:1349998059319:dw| so i multiply the exponent to the 3rd?

  18. pokemon23
    • 3 years ago
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    i know its wrong..

  19. Calcmathlete
    • 3 years ago
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    If you were to actually multiply it out, it would be like doing this: \[(9x + 6)^3 \implies (9x + 6)(9x + 6)(9x + 6)\]which is unnecessary here. Again, like previous times, do you see anything that you can factor out?

  20. pokemon23
    • 3 years ago
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    i keep forgetting to factor.....

  21. pokemon23
    • 3 years ago
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    my spectacles are deceiving me

  22. Calcmathlete
    • 3 years ago
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    Hmm?

  23. pokemon23
    • 3 years ago
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    ok if we can (9x+6) (9x+6) (9x+6)

  24. pokemon23
    • 3 years ago
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    what's next?

  25. Calcmathlete
    • 3 years ago
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    THat is not a step that we even need to take, but we can solve it using that. It would be a slightly longer process though. Let's work WITHIN the parentheses first. Don't do anything ith the \(^3\) for now.

  26. pokemon23
    • 3 years ago
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    ok will keep it (9x+6)^3

  27. Calcmathlete
    • 3 years ago
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    Now factor what's within the parentheses...

  28. pokemon23
    • 3 years ago
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    3(3x+2)^3

  29. Calcmathlete
    • 3 years ago
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    The 3 wouldn't go outside of the parentheses because it's still under the \(^3\), so it would become \((3(3x + 2))^3\) Do you see why that is?

  30. pokemon23
    • 3 years ago
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    explain those parentheses

  31. Calcmathlete
    • 3 years ago
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    Well, let's take the approach that you had before. \[(9x + 6)^3 \implies (9x + 6)(9x + 6)(9x + 6) \implies 3(3x + 2)3(3x + 2)3(3x + 2)\]\[3 \times 3 \times 3(3x + 2)(3x + 2)(3x + 2) \implies 3^3(3x + 2)^3~~~OR~~~27(3x + 2)^3\]See what I did? You can't take the 3 \(\LARGE{ENTIRELY}\) out of the parentheses unless you take the \(^3\) with it.

  32. pokemon23
    • 3 years ago
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    oh

  33. Calcmathlete
    • 3 years ago
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    \[\LARGE{¿Entiendes?}\]

  34. pokemon23
    • 3 years ago
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    si

  35. Calcmathlete
    • 3 years ago
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    Can you predict what to do next?

  36. pokemon23
    • 3 years ago
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    distributive

  37. Calcmathlete
    • 3 years ago
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    What? Keep in mind the original question and what y is equal to...

  38. pokemon23
    • 3 years ago
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    then we should substitute

  39. Calcmathlete
    • 3 years ago
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    \[\huge\color{red}{Y}\color{blue}{E}\color{salmon}{S}\color{green}{!}\]

  40. pokemon23
    • 3 years ago
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    so ....

  41. pokemon23
    • 3 years ago
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    instead of the x make it a y?

  42. Calcmathlete
    • 3 years ago
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    Uh...remember that \(y = (3x + 2)^3...\) You can substitute y in for that value now. Tell me what your final answer would be now.

  43. pokemon23
    • 3 years ago
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    jinkies

  44. pokemon23
    • 3 years ago
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    hmm ummm

  45. pokemon23
    • 3 years ago
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    y=27x^3+8

  46. Calcmathlete
    • 3 years ago
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    Umm...no... \[(9x + 6)^3 \implies (3(3x + 2))^3 \implies 3^3(3x + 2)^3 \implies 27(3x + 2)^3 \implies 27y\]

  47. Calcmathlete
    • 3 years ago
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    Do you see why? Do you understand what it means when it says "in terms of y"? Do you see that we're replacing ALL OF \((3x + 2)^3\) with y? That way nothing in that is cubed anymore?

  48. pokemon23
    • 3 years ago
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    I still don't get how divided 27(3x+2)^3 to get 27y

  49. Calcmathlete
    • 3 years ago
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    Where do you see a division symbol? There is no division invloved in this except the beginning factoring portion. \[\large{\text{We just clarified that y}}~\LARGE\text{=}~\large{(3x + 2)^3}\]Just replace \((3x + 2)^3\) with y and that's all that happened.

  50. pokemon23
    • 3 years ago
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    where does (3x+2)^3? disappear?

  51. pokemon23
    • 3 years ago
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    I don't understand the y part...

  52. Calcmathlete
    • 3 years ago
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    fdiofvejbvnweokgnweoifnerwo That is what we're substituting with y! Remember when we said y = 2 and we plugged it in? Well we could say that 2 = y and we plug y in for the 2. SAME THING HERE. y = (3x + 2)^3 (3x + 2)^3 = y Now you just plug in why for the (3x + 2)^3

  53. Calcmathlete
    • 3 years ago
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    You still there?

  54. pokemon23
    • 3 years ago
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    im here

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