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anonymous

  • 5 years ago

Question can someone perform this operation? simplify the expression: numerator (6r^2-7r-3) divided by denominator (r^2-1) divided by numerator (4r^2-12r+9) divided by denominator (r^2-1) multiply by numerator (2r^2-r-3) divided by denominator (3r^2-2r-1)

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  1. anonymous
    • 5 years ago
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    Can I just give the answer out? Too much typing.

  2. anonymous
    • 5 years ago
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    Yes

  3. anonymous
    • 5 years ago
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    Your question is worded a little bit strangely... When you say numerator and then the denominator, do you mean each of those are one set of terms? aka (6r^2 - 7r - 3) / (r^2 - 1) ------------------------- <--divided by (4r^2 - 12r + 9) / (r^2 - 1) ------------------------- (2r^2 - r - 3) / (3r^2 - 2r - 1)

  4. anonymous
    • 5 years ago
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    Yes like numerator 12 divided by denominator 2=6

  5. amistre64
    • 5 years ago
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    Can you write this out using grouping symbols to make the question easier to read?

  6. anonymous
    • 5 years ago
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    Yes all of the numerators are numerator but in sets like you have up top there.

  7. amistre64
    • 5 years ago
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    For example: 4/2/2 can be taken several ways: (4/2)/2 = 2/2 = 1 4/(2/2) = 4/1 = 4 As you can see; 4 does not equal 1 so the way the problem is set up is important

  8. anonymous
    • 5 years ago
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    Sorry, going to start it now.

  9. anonymous
    • 5 years ago
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    (3r + 1)^ 2 * (r^2 - 1)

  10. amistre64
    • 5 years ago
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    If you define a way to set the problem up like using "//" to be the determining division between top and bottom... that would be helpful: example: 4//2/2 = 4 4/2//2 = 1

  11. anonymous
    • 5 years ago
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    (r+1)/(r-1)

  12. anonymous
    • 5 years ago
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    6r^2-7r-3/r^2-1 / 4r^2-12r+9/r^2-1 times 2r^2-r-3/3r^2-2r-1

  13. anonymous
    • 5 years ago
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    Beens that looks like the right solution, does anyone have more input?

  14. amistre64
    • 5 years ago
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    (2r-3)(3r+1) -------- (r+1)(r-1) (2r-3)(r+1) -------------- times --------- = ? (2r-3)(2r-3) (r-1)(3r+1) ---------- (r+1)(r-1)

  15. amistre64
    • 5 years ago
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    If so, I get: (3r+1)(2r-3)(r+1) (r+1) --------------- = ----- same as Beens (2r-3)(r-1)(3r+1) (r-1)

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