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Math Help - Help!! Complex Fractions

  1. #1
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    Help!! Complex Fractions

    can anybody help me?! I am really confused

    1.) a-(5a/a+5) / a+(5a/a-5)

    2.) (a/a-b) - (b/a+b) / (b/a-b) + (a/a+b)

    3.) (a-b/a+b) + (b/a) / (a/b) - (a-b/a+b)



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  2. #2
    Senior Member OReilly's Avatar
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    Quote Originally Posted by whoa View Post
    can anybody help me?! I am really confused

    1.) a-(5a/a+5) / a+(5a/a-5)

    2.) (a/a-b) - (b/a+b) / (b/a-b) + (a/a+b)

    3.) (a-b/a+b) + (b/a) / (a/b) - (a-b/a+b)



    Thanks
    1) \frac{{a - \frac{{5a}}{{a + 5}}}}{{a + \frac{{5a}}{{a - 5}}}} = \frac{{\frac{{a(a + 5) - 5a}}{{a + 5}}}}{{\frac{{a(a - 5) + 5a}}{{a - 5}}}} = \frac{{\frac{{a^2  + 5a - 5a}}{{a + 5}}}}{{\frac{{a^2  - 5a + 5a}}{{a - 5}}}} = \frac{{\frac{{a^2 }}{{a + 5}}}}{{\frac{{a^2 }}{{a - 5}}}} = \frac{{a^2 (a - 5)}}{{a^2 (a + 5)}} = \frac{{a - 5}}{{a + 5}}

    2) \frac{{\frac{a}{{a - b}} - \frac{b}{{a + b}}}}{{\frac{b}{{a - b}} + \frac{a}{{a + b}}}} = \frac{{\frac{{a(a + b) - b(a - b)}}{{(a - b)(a + b)}}}}{{\frac{{b(a + b) + a(a - b)}}{{(a - b)(a + b)}}}} = \frac{{\frac{{a^2  + ab - ab + b^2 }}{{a^2  - b^2 }}}}{{\frac{{ab + b^2  + a^2  - ab}}{{a^2  - b^2 }}}} = \frac{{\frac{{a^2  + b^2 }}{{a^2  - b^2 }}}}{{\frac{{a^2  + b^2 }}{{a^2  - b^2 }}}} = \frac{{(a^2  + b^2 )(a^2  - b^2 )}}{{(a^2  - b^2 )(a^2  + b^2 )}} = 1

    3) \frac{{\frac{{a - b}}{{a + b}} + \frac{b}{a}}}{{\frac{a}{b} - \frac{{a - b}}{{a + b}}}} = \frac{{\frac{{a(a - b) + b(a + b)}}{{a(a + b)}}}}{{\frac{{a(a + b) - b(a - b)}}{{b(a + b)}}}} = \frac{{\frac{{a^2  - ab + ab + b^2 }}{{a(a + b)}}}}{{\frac{{a^2  + ab - ab + b^2 }}{{b(a + b)}}}} = \frac{{\frac{{a^2  + b^2 }}{{a(a + b)}}}}{{\frac{{a^2  + b^2 }}{{b(a + b)}}}} = \frac{{(a^2  + b^2 )b(a + b)}}{{(a^2  + b^2 )a(a + b)}} = \frac{b}{a}
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  3. #3
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    thanks! by the way how did you get the final answer?
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  4. #4
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    Hello, whoa!

    The standard procedure for complex fractions is to multiply through by the LCD.
    This allows us to cancel those annoying denominators.


    1)\;\frac{a - \frac{5a}{a+5}}{a + \frac{5a}{a-5}}

    The LCD is (a+5)(a-5)

    . . \frac{(a+5)(a-5)\cdot\left[a - \frac{5a}{a+5}\right]} {(a+5)(a-5)\cdot\left[a + \frac{5a}{a-5}\right] }\;= \;\frac{a(a+5)(a-5) - 5a(a-5)}{a(a+5)(a-5) + 5a(a+5)}

    Factor: . \frac{a\cdot(a-5)\cdot[(a+5) - 5]}{a\cdot(a+5)\cdot[(a-5) + 5]} \;=\;\frac{a\cdot(a-5)\cdot a}{a\cdot(a+5)\cdot a} \;= \;\boxed{\frac{a-5}{a+5}}



    2)\;\frac{\frac{a}{a-b} - \frac{b}{a+b}}{\frac{b}{a-b} + \frac{a}{a+b}}

    The LCD is (a- b)(a+b)

    . . \frac{(a-b)(a+b)\cdot\left[\frac{a}{a-b} - \frac{b}{a+b}\right]}{(a-b)(a+b)\cdot\left[\frac{b}{a-b} + \frac{a}{a+b}\right]} \;= \;\frac{a(a+b) - b(a-b)}{b(a+b) + a(a-b)}

    . . . = \;\frac{a^2+ab - ab + b^2}{ab + b^2 + a^2 - ab} \;=\;\frac{a^2+b^2}{a^2+b^2} \;= \;\boxed{1}



    3)\;\frac{\frac{a-b}{a+b} + \frac{b}{a}}{\frac{a}{b} - \frac{a-b}{a+b}}

    The LCD is ab(a+b)

    . . \frac{ab(a+b)\cdot\left[\frac{a-b}{a+b} + \frac{b}{a}\right]}{ab(a+b)\cdot\left[\frac{a}{b} - \frac{a-b}{a+b}\right]}<br />
\;= \;\frac{ab(a-b) + b^2(a + b)}{a^2(a+b) - ab(a-b)} = \;\frac{a^2b - ab^2 + ab^2 + b^3}{a^3 + a^2b - a^2b + ab^2}

    . . = \;\frac{a^2b + b^3}{a^3 + ab^2} \;=\;\frac{b(a^2+b^2)}{a(a^2+b^2)} \;=\;\boxed{\frac{b}{a}}

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  5. #5
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    Ah ok. thanks............ but I still have a problem. what if the numbers are raised by negative one? here are some examples:

    1.) (3x^-1 - 2y^-1) / (2y^-1 - 3x^-1)

    2.) {(3a)^-1 - (2b)^-1} / {(2b)^-1 + (3a)^-1}

    3.) (ab^-1 + ba^-1) / {(ab)^-1}

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  6. #6
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    Quote Originally Posted by whoa View Post
    Ah ok. thanks............ but I still have a problem. what if the numbers are raised by negative one? here are some examples:

    1.) (3x^-1 - 2y^-1) / (2y^-1 - 3x^-1)

    2.) {(3a)^-1 - (2b)^-1} / {(2b)^-1 + (3a)^-1}

    3.) (ab^-1 + ba^-1) / {(ab)^-1}

    General rule when having number raised by negative number is:

    \frac{1}{{a^m }} = a^{ - m} ,a \in R\backslash \{ 0\} ,m \in N

    I will show you examples from your equations for such numbers but you should try to do all this by yourself since you have been given help from Soroban and me for similar examples.

    So we have 3x^{ - 1}  = 3\frac{1}{x} = \frac{3}{x}
    Notice that we only raised x by -1 not 3x.

    You have also (3a)^{ - 1}  = \frac{1}{{3a}}.
    Notice that we raised 3a by -1 not just a.

    Try to do something and this forum will help you in your effort.
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  7. #7
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    uhm... i'm not sure here's my answer:

    1.) (3/x -2/y)/(2/y-3/x) = {3(y)-2(x)/xy} = (3y-2x/xy)/(2x-3y/xy)= (3y-2x/xy) * (xy/2x-3y)=////////// (3y-2x)/(2x-3y)final answer

    2.) (1/3a-1/2b)/(1/2b+1/3a) = {2b-3a/(3a)(2b)}/{(3a+2b/(2b)(3a)}={2b-3a/(3a)(2b)}*{3a+2b/(2b)(3a)}= ////////-3a/3a,,, final answer

    3.) (1/ab+1/ba)/(1/ab)= (1/ab)/(1/ab)= (1/ab*ab/1)=///////////1,,, final answer

    hehe i'm not sure
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  8. #8
    Senior Member OReilly's Avatar
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    Quote Originally Posted by whoa View Post
    uhm... i'm not sure here's my answer:

    1.) (3/x -2/y)/(2/y-3/x) = {3(y)-2(x)/xy} = (3y-2x/xy)/(2x-3y/xy)= (3y-2x/xy) * (xy/2x-3y)=////////// (3y-2x)/(2x-3y)final answer

    2.) (1/3a-1/2b)/(1/2b+1/3a) = {2b-3a/(3a)(2b)}/{(3a+2b/(2b)(3a)}={2b-3a/(3a)(2b)}*{3a+2b/(2b)(3a)}= ////////-3a/3a,,, final answer

    3.) (1/ab+1/ba)/(1/ab)= (1/ab)/(1/ab)= (1/ab*ab/1)=///////////1,,, final answer

    hehe i'm not sure
    1) Correct.

    2) Answer is not correct. You have done it almost ok but at the end you made an error.
    Solution is:
    \frac{{\frac{1}{{3a}} - \frac{1}{{2b}}}}{{\frac{1}{{2b}} + \frac{1}{{3a}}}} = \frac{{\frac{{2b - 3a}}{{6ab}}}}{{\frac{{3a + 2b}}{{6ab}}}} = \frac{{6ab(2b - 3a)}}{{6ab(3a + 2b)}} = \frac{{2b - 3a}}{{3a + 2b}}

    \frac{{2b - 3a}}{{3a + 2b}} can't be further simplifed as you did to  - \frac{{3a}}{{3a}}

    3) Not correct. You made a mistake at start.

    It's not ab^{ - 1}  = \frac{1}{{ab}} because only b is raised by -1. If it were (ab)^{ - 1} then it would be (ab)^{ - 1}  = \frac{1}{{ab}}.
    Correct is ab^{ - 1}  = a\frac{1}{b} = \frac{a}{b}.

    Now, try to do 3) again.
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  9. #9
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    wow! thanks

    here's no. 3 i'm not sure again hehe

    (a/b+b/a) / (1/ab)= [a(b)+b(a)/ab] / 1/ab= (ab+ba/ab) / 1/ab= (ab+ba/ab)*ab/1=///////// (ab+ba/1),,,, final answer
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  10. #10
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    Hello, whoa!

    Why don't you try it my way? . . . Or don't you understand it?


    1)\;\;\frac{3x^{-1} - 2y^{-1}}{2y^{-1} - 3x^{-1}}

    We have: . \frac{\frac{3}{x} - \frac{2}{y}}{\frac{2}{y} - \frac{3}{x}} . . . . The LCD is xy

    . . \frac{xy\,\left(\frac{3}{x} - \frac{2}{y}\right)}{xy\,\left(\frac{2}{y} - \frac{3}{x}\right)} \;=\;\frac{\not{x }y\cdot\frac{3}{\not{x }} - x\!\!\not{y }\cdot\frac{2}{\not{y }}}{x\!\!\not{y}\cdot\frac{2}{\not{y}} - \not{x}y\cdot\frac{3}{\not{x}}} \;=\;\frac{3y - 2x}{2x - 3y} \:=\:\frac{-(2x - 3y)}{2x-3y} \;=\;-1



    2)\;\;\frac{(3a)^{-1} - (2b)^{-1}}{2b)^{-1} + (3a)^{-1}}

    We have: . \frac{\frac{1}{3a} - \frac{1}{2b}}{\frac{1}{2b} + \frac{1}{3a}} . . . . . The LCD is 6ab

    . . \frac{6ab\left(\frac{1}{3a} - \frac{1}{2b}\right)}{6ab\left(\frac{1}{2b} + \frac{1}{3a}\right)} \;= \;\frac{\not{6 }^2\!\!\not{a}b\cdot\frac{1}{\not{3}\not{a}} - \not{6 }^3a\!\!\not{b}\cdot\frac{1}{\not{2}\not{b}}}{\not  {6}^3a\!\!\not{b}\cdot\frac{1}{\not{2}\not{b}} - \not{6}^2\!\!\not{a}b\cdot\frac{1}{\not{3}\not{a}}  } \;=\;\frac{2b -3a}{3a+2b} . . .
    cannot be reduced


    3)\;\;\frac{ab^{-1} + ba^{-1}}{(ab)^{-1}}

    We have: . \frac{\frac{a}{b} + \frac{b}{a}}{\frac{1}{ab}} . . . .The LCD is ab

    . . \frac{ab\left(\frac{a}{b} + \frac{b}{a}\right)}{ab\left(\frac{1}{ab}\right)} \;= \;\frac{a\!\!\not{b}\cdot\frac{a}{\not{b}} + \not{a }b\cdot\frac{b}{\not{a }}}{\not{a }\!\!\not{b}\cdot\frac{1}{\not{a}\not{b}}} \;=\;\frac{a^2+b^2}{1}\;=\;a^2+b^2

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  11. #11
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    thanks for lending your time to me you guys I'd really appreciate it! I learned a lot!

    @ soroban

    . . \frac{6ab\left(\frac{1}{3a} - \frac{1}{2b}\right)}{6ab\left(\frac{1}{2b} + \frac{1}{3a}\right)} \;= \;\frac{\not{6 }^2\!\!\not{a}b\cdot\frac{1}{\not{3}\not{a}} - \not{6 }^3a\!\!\not{b}\cdot\frac{1}{\not{2}\not{b}}}{\not  {6}^3a\!\!\not{b}\cdot\frac{1}{\not{2}\not{b}} - \not{6}^2\!\!\not{a}b\cdot\frac{1}{\not{3}\not{a}}  } \;=\;\frac{2b -3a}{3a+2b} . . . [/size]cannot be reduced
    why cant it be reduced?






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  12. #12
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    ok know it already! thanks anyways
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