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Math Help - Rational Expressions and Radical Expressions

  1. #1
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    Rational Expressions and Radical Expressions

    I just want to check my answers with someone because im not sure i am correct...

    1. 12u cubed/7v to the fifth * 14v squared/3u =
    8 u squared/ v cubed

    2. 9x * 4/x =
    36?

    3. x squared -3x +2/x squared -4x +4 divided by x squared -2x +1/ 3x squared -12 =

    3(x+2)/ (x-1)

    4. 15 divided by 5x/x+1 =
    3(x+1) / x

    5. 6x/x-2 divided by 3x =
    2/x-2
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  2. #2
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    You really should learn to type in LaTex. I can barely read your post.

    1. 12u cubed/7v to the fifth * 14v squared/3u =
    8 u squared/ v cubed
    You mean this?
    \frac{12u^3}{7v^5} \cdot \frac{14v^2}{3u}
    If so, then
    \frac{12u^3}{7v^5} \cdot \frac{14v^2}{3u} = \frac{4u^2}{v^3} \cdot \frac{2}{1} = \frac{8u^2}{v^3}
    It's correct.

    2. 9x * 4/x =
    36?
    You mean this?
    9x \cdot \frac{4}{x}
    If so, and assuming that x isn't 0, then 36 is right.

    3. x squared -3x +2/x squared -4x +4 divided by x squared -2x +1/ 3x squared -12 =

    3(x+2)/ (x-1)
    I'm not even touching this one.

    4. 15 divided by 5x/x+1 =
    3(x+1) / x
    You mean this?
    15 \div \frac{5x}{x + 1}
    = 15 \cdot \frac{x + 1}{5x} = \frac{3(x + 1)}{x}

    5. 6x/x-2 divided by 3x =
    2/x-2
    You mean this?
    \frac{6x}{x - 2} \div 3x
    = \frac{6x}{x - 2} \cdot \frac{1}{3x} = \frac{2}{x - 2}


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  3. #3
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    Thanks again,
    how would you solve problem involving addition???

    ex) x/ (x squared-9) + 3/(x squared-9)
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  4. #4
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    Quote Originally Posted by illeatyourxface View Post
    I just want to check my answers with someone because im not sure i am correct...

    1. 12u cubed/7v to the fifth * 14v squared/3u =
    8 u squared/ v cubed

     \frac{12u^3}{7v^5}\times\frac{14v^2}{3u}

     =\frac{12u^3\times 14v^2}{7v^5\times 3u}

     =\frac{12u^3\times 14v^2}{7v^5\times 3u}

     =\frac{8u^3\times v^2}{v^5\times u}

     =8u^2\times v^{-3}

     =\frac{8u^2}{ v^{-3}}
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  5. #5
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    Quote Originally Posted by illeatyourxface View Post
    Thanks again,
    how would you solve problem involving addition???

    ex) x/ (x squared-9) + 3/(x squared-9)
    If the denominators are the same, just add the numerators up, like this:
    \frac{x}{x^2 - 9} + \frac{3}{x^2 - 9} = \frac{x + 3}{x^2 - 9}

    But we're not done, because the denominator can be factored:
    \frac{x + 3}{x^2 - 9} = \frac{x + 3}{(x + 3)(x - 3)}

    Cancel out the x + 3, and the answer is
    \frac{1}{x - 3}


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  6. #6
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    Quote Originally Posted by pickslides View Post
     \frac{12u^3}{7v^5}\times\frac{14v^2}{3u}

     =\frac{12u^3\times 14v^2}{7v^5\times 3u}

     =\frac{12u^3\times 14v^2}{7v^5\times 3u}

     =\frac{8u^3\times v^2}{v^5\times u}

     =8u^2\times v^{-3}

     =\frac{8u^2}{ v^{-3}}


    I thought negative exponents were used only if you moved it from the top or bottom...??
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  7. #7
    Senior Member Stroodle's Avatar
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    Quote Originally Posted by illeatyourxface View Post

    3. x squared -3x +2/x squared -4x +4 divided by x squared -2x +1/ 3x squared -12 =
    Do you mean:

    \frac{x^2-3x+2}{x^2-4x+4}\div\frac{x^2-2x+1}{3x^2-12}

    =\frac{(x-2)(x-1)}{(x-2)^2}\div\frac{(x-1)^2}{3(x+2)(x-2)}

    =\frac{3(x-2)^2(x+2)(x-1)}{(x-2)^2(x-1)^2}

    =\frac{3(x+2)}{x-1}
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  8. #8
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    okayyy how would you add or subtract if the demoninator was different?


    ex)

    4/x cubed y + 7/x y squared
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  9. #9
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    and im really sorry if this is confusing i dont know how to type in LAtex, im sorrry
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  10. #10
    Senior Member Stroodle's Avatar
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    \frac{12u^3}{7v^5}\times\frac{14v^2}{3u}

    =\frac{168u^3v^2}{21v^5u}

    =\frac{8u^2}{v^3}

    edit* Oops I didn't notice that Yeongil already answered this one correctly...
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  11. #11
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    Quote Originally Posted by illeatyourxface View Post
    I thought negative exponents were used only if you moved it from the top or bottom...??
    yep my bad, forgot to remove the "-" sign
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  12. #12
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    to solve it.... would it be like

    4/ x cubed y + 7/ x y squared =


    (4)(y) + (7) (x squared) / (x cubed y)(x y squared) ????
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  13. #13
    Senior Member Stroodle's Avatar
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    Do you mean:

    \frac{4}{x^3y}+\frac{7}{xy^2}

    =\frac{4y}{x^3y^2}+\frac{7x^2}{x^3y^2}

    =\frac{4y+7x^2}{x^3y^2}
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  14. #14
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    okay soo if you are subtracting...

    11/3x - 5/6x =

    33x -5 / 6x ?
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  15. #15
    Senior Member Stroodle's Avatar
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    \frac{11}{3x}-\frac{5}{6x}

    To make the denominator the same for both terms you multiply the top and bottom of the first term by 2

    =\frac{22}{6x}-\frac{5}{6x}

    =\frac{22-5}{6x}

    =\frac{17}{6x}
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