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anonymous

  • 5 years ago

5x(3x+2)-6 < 4X(2X-5) +12

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  1. anonymous
    • 5 years ago
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    Hello

  2. anonymous
    • 5 years ago
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    Do you have any guesses on how to start this problem?

  3. anonymous
    • 5 years ago
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    I multipy 5x 15+10-6< 4x 12x-20+12, I am not good at all with algebra.

  4. anonymous
    • 5 years ago
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    You were indeed correct to start off by distributing. But we have a couple minor errors with your distribution, however.

  5. anonymous
    • 5 years ago
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    When you multiply 5x times 3x, what do you get?

  6. anonymous
    • 5 years ago
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    15

  7. anonymous
    • 5 years ago
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    Actually, you get 15x^2. The x's multiply with each other. But the 5*3 was spot on.

  8. anonymous
    • 5 years ago
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    okay, thanks!

  9. anonymous
    • 5 years ago
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    Do you see how that works?

  10. anonymous
    • 5 years ago
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    Do you see how that works?

  11. anonymous
    • 5 years ago
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    Alright, now let's try to multiply the 5x and the +2 together.

  12. anonymous
    • 5 years ago
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    5x + 2 -7

  13. anonymous
    • 5 years ago
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    5x(3x+2)-6 Hmm.. Not quite. Maybe you don't know where I am talking about. In the above phrase, will you take the 5x and multiply it with the 2 in the parenthesis?

  14. anonymous
    • 5 years ago
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    Yes, 5x2+10

  15. anonymous
    • 5 years ago
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    Mm.. Not quite. Remember, the x is a variable, not the same as a multiplication symbol. 5x * 2 = 10x

  16. anonymous
    • 5 years ago
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    Oh, and by the way, * is the computer symbol for multiplication. And that's the reason why, we don't want to get confused about whether it's a letter x or an operation so we use *.

  17. anonymous
    • 5 years ago
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    Okay I understand

  18. anonymous
    • 5 years ago
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    Great. So if you want to double (or multiply by two) your five X's, you now have 10 X's. Is that clear?

  19. anonymous
    • 5 years ago
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    Yes!

  20. anonymous
    • 5 years ago
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    So on the left side we went from: 5x(3x+2)-6 To: 15x^2 + 10x - 6 Following? Now we'll want to do distribute on the right side: 4X(2X-5) +12

  21. anonymous
    • 5 years ago
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    Okay I am following!

  22. anonymous
    • 5 years ago
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    Cool, can you distribute on the right side then? 4X(2X-5) +12

  23. anonymous
    • 5 years ago
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    yes 20x^4 +10

  24. anonymous
    • 5 years ago
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    Hmm.. Let's try this together. 4X(2X) Can you do this for me? Remember what I said above about multiply two Xs together.

  25. anonymous
    • 5 years ago
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    8x^2 +12

  26. anonymous
    • 5 years ago
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    Excellent, very very good. Now you left out one last part for the right side: 4X(2X-5) +12 Can you do the 4x * -5?

  27. anonymous
    • 5 years ago
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    4x8-5+3

  28. anonymous
    • 5 years ago
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    I'm afraid not. 4x * -5 = -20x Pretend that an X is an item. Don't even bother calling it a variable. Now, I have 4 Xs. If I have 4 Xs and I multiply my Xs by negative five, how many Xs do I have?

  29. anonymous
    • 5 years ago
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    would it be 1x

  30. anonymous
    • 5 years ago
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    Hmm.. Let's try something a little different. If I have 4 pennies and you said that you could multiply my pennies by five, by the end of it, how many pennies would I have?

  31. anonymous
    • 5 years ago
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    20

  32. anonymous
    • 5 years ago
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    Exactly. Now, pretend that my pennies were really just little X's all along. Not a letter, not a variable, a real X, something I could hold in my hands. I'll ask you the same question. If I have 4 X's in my hand and you said you could multiply my X's by 5, how many X's would I have?

  33. anonymous
    • 5 years ago
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    20x's

  34. anonymous
    • 5 years ago
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    Great, does that make it a little more clear? That is how you should approach these problems when multiplying X's. Just pretend they are just labels. :)

  35. anonymous
    • 5 years ago
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    Yes it does, thank you!

  36. anonymous
    • 5 years ago
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    So we had this: 5x(3x+2)-6 < 4X(2X-5) +12 And turned it into this: 15x^2 + 10x - 6 < 20x^2 - 20x + 12 When we solve algebraically, we'll want everything that has that nasty X on the same side of the inequality. Do you know how you might accomplish that?

  37. anonymous
    • 5 years ago
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    15x2+10x-6 < 20x2-20x+12 -15x+6 20x-12

  38. anonymous
    • 5 years ago
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    Afraid not. We'll want to do: 15x^2 + 10x - 6 < 20x^2 - 20x + 12 +6 -20x^2 +20x +6 To get: 15x^2 +10x - 20x^2 + 20x < 18

  39. anonymous
    • 5 years ago
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    From there we can combine like terms: The x^2s can combine. 15x^2 - 20x^2 = -5x^2 The X's can combine. 10x + 20x = 30x So we end up with: -5x^2 + 30x < 18.

  40. anonymous
    • 5 years ago
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    You subtracted 2 on both sides to get the answer 18 and combined like terms 15x 20x and 5x and added 10+20 = 30.

  41. anonymous
    • 5 years ago
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    I added six to both sides to get +18 on the right side. The rest is correct :)

  42. anonymous
    • 5 years ago
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    Now, are you familiar with the quadratic equation?

  43. anonymous
    • 5 years ago
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    Yes I understand somewhat, however I will continue to practice and study for a better understanding. Thank you so much for your help!

  44. anonymous
    • 5 years ago
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    It's no problem. I just wish I could be a little more clear, it's hard to do that with just text. Anyways, we'll use the quadratic equation, first we have to make the equation equal to 0: -5x^2 + 30x < 18 -18 -18 -5x^2 + 30x -18 < 0 And then plug in the A, B, and C values into the quadratic equation: -b plus/minus sqrt of b^2 - 4AC all divided by 2A - 30 plus/minus sqrt 30^2 - 4*-5*-18, all divided by 2(-5) = 16.905 and 133.09. Those are your values of X and those are the answers. Two of them.

  45. anonymous
    • 5 years ago
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    Oh wait. I lie.

  46. anonymous
    • 5 years ago
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    Okay

  47. anonymous
    • 5 years ago
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    I screwed up my math somewhere. The system is true when: -4.819 < X < 0.533

  48. anonymous
    • 5 years ago
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    And that's the answer.

  49. anonymous
    • 5 years ago
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    Okay thank you!!!! I am going to print out this information and study it and I will practice other problems. Thanks again!!!!

  50. anonymous
    • 5 years ago
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    No problem Sorry I couldn't be more clear.

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