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Math Help - Help solve, I got quiz tomorow on this stuff need quick help!

  1. #1
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    Help solve, I got quiz tomorow on this stuff need quick help!

    Allright so tomorow I got a quiz on this stuff. I just need someone to show me step by step how to do this!

     <br />
\frac{bx}{a}+\frac{1}{5}=2x SOLVE FOR X


     <br />
\frac{3}{M}=\frac{M}{5} SOLVE FOR M


     <br />
\frac{1}{U}=\frac{y}{k}+\frac{1}{h} SOLVE FOR K


     <br />
\frac{V-6}{5}+\frac{V-8}{15}+\frac{V}{10}=0 SOLVE FOR V


     <br />
\frac{5}{w+3}+4=15 SOLVE FOR w


    and some similar problems that I need to isolate

     <br />
\frac{K}{M}=P+\frac{a}{M^2} ISOLATE a


    Thanks!!!
    Last edited by elitescouter; July 8th 2009 at 10:40 PM.
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  2. #2
    Senior Member Stroodle's Avatar
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    \frac{bx}{a}+\frac{1}{5}=2x

    First establish a common denominator:

    \frac{5bx}{5a}+\frac{a}{5a}=\frac{10ax}{5a}

    Then solve for x

    10ax-5bx=a

    x(10a-5b)=a

    x=\frac{a}{10a-5b}

    x=\frac{a}{5(2a-b)}

    Use the same process for this one:

    \frac{1}{U}=\frac{y}{k}+\frac{1}{h}

    \frac{hk}{hkU}=\frac{hUy}{hkU}+\frac{kU}{hkU}

    hk-kU=hUy

    k(h-U)=hUy

    \therefore k=\frac{hUy}{h-U}

    Can you now use this process to do the other ones? Each term just needs to have the same denominator.



    For the last question:

    \frac{K}{M}=P+\frac{a}{M^2}

    \frac{a}{M^2}=\frac{K}{M}-P

    a=M^2(\frac{K}{M}-P)

    a=\frac{KM^2}{M}-PM^2

    a=KM-PM^2

    a=M(K-PM)
    Last edited by Stroodle; July 8th 2009 at 11:11 PM.
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  3. #3
    Senior Member Stroodle's Avatar
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    \frac{5}{w+3}+4=15

    For this question you can multiply each term by (w+3) to get rid of the fraction.

    5+4(w+3)=15(w+3)

    4w+17=15w+45

    11w=-28

    \therefore w=-\frac{28}{11}
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  4. #4
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    Sorry to bother u but u think u can help me with

    img.top {vertical-align:15%;}  <br />
\frac{3}{M}=\frac{M}{5} SOLVE FOR M" alt="  <br />
\frac{3}{M}=\frac{M}{5} SOLVE FOR M" />

    I get stuck half way through it.
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  5. #5
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    Quote Originally Posted by elitescouter View Post
    Sorry to bother u but u think u can help me with

    \frac{3}{M}=\frac{M}{5} SOLVE FOR M

    I get stuck half way through it.
    Just cross multiply:
    \begin{aligned}<br />
\frac{3}{M} &= \frac{M}{5} \\<br />
M^2 &= 15<br />
\end{aligned}

    Now find M.


    01
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  6. #6
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    Quote Originally Posted by elitescouter View Post
    Sorry to bother u but u think u can help me with

    img.top {vertical-align:15%;}  <br />
\frac{3}{M}=\frac{M}{5} SOLVE FOR M" alt="  <br />
\frac{3}{M}=\frac{M}{5} SOLVE FOR M" />

    I get stuck half way through it.
    make every term have the same denominator.
    M^2=3*5
    M=sqrt (15)
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  7. #7
    Senior Member Stroodle's Avatar
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    Don't forget the \pm

    m^2=15

    \therefore m=\pm \sqrt{15}
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  8. #8
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    oops forgot that!
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  9. #9
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    Quote Originally Posted by yeongil View Post
    Just cross multiply:
    \begin{aligned}<br />
\frac{3}{M} &= \frac{M}{5} \\<br />
M^2 &= 15<br />
\end{aligned}

    Now find M.


    01
    What other alternatives other than cross multiplying are possible here? Because if it were me I would multiply by the reciprocal \frac{m}{3}*\frac{3}{m}=\frac{m}{5}*\frac{m}{3}. Which would equal \frac{m^2}{15}
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  10. #10
    Senior Member Stroodle's Avatar
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    You can use a common denominator:

    \frac{3}{M}=\frac{M}{5}

    \frac{15}{5M}=\frac{M^2}{5M}

    M^2=15

    \therefore M=\pm\sqrt{15}
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  11. #11
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    Quote Originally Posted by allyourbass2212 View Post
    What other alternatives other than cross multiplying are possible here? Because if it were me I would multiply by the reciprocal \frac{m}{3}*\frac{3}{m}=\frac{m}{5}*\frac{m}{3}. Which would equal \frac{m^2}{15}
    Your method would be fine as long as you keep your eye on BOTH sides of the equation.

    \frac{3}{M} = \frac{M}{5}

    \frac{3}{M}\cdot\frac{M}{3} = \frac{M}{5}\cdot\frac{M}{3}

    1 = \frac{M^2}{15}

    15 = M^2

    \pm\sqrt{15} = M.
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  12. #12
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    Quote Originally Posted by Stroodle View Post
    \frac{5}{w+3}+4=15

    For this question you can multiply each term by (w+3) to get rid of the fraction.

    5+4(w+3)=15(w+3)

    4w+17=15w+45

    11w=-28

    \therefore w=-\frac{28}{11}
    It's easier if you subtract 4 from both sides first...

    \frac{5}{w + 3} + 4 = 15

    \frac{5}{w + 3} = 11

    \frac{w + 3}{5} = \frac{1}{11}

    w + 3 = \frac{5}{11}

    w = \frac{5}{11} - 3

    w = -\frac{28}{11}.
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  13. #13
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    Embarrassingly enough the basic algebra book I am reading through does not cover how to solve when the equation looks as such.

    1 = \frac{M^2}{15}

    15 = M^2

    Would you please explain the steps involved in how 1 = \frac{M^2}{15} equals 15 = M^2
    ?

    Many thanks
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  14. #14
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    Quote Originally Posted by allyourbass2212 View Post
    Embarrassingly enough the basic algebra book I am reading through does not cover how to solve when the equation looks as such.

    1 = \frac{M^2}{15}

    15 = M^2

    Would you please explain the steps involved in how 1 = \frac{M^2}{15} equals 15 = M^2
    ?

    Many thanks
    Can you see that M^2 is divided by 15?

    To undo this, we have to multiply both sides by 15.


    So 1 = \frac{M^2}{15}

    1\cdot 15 = \frac{M^2}{15}\cdot 15

    15 = M^2


    Then to undo the squaring, we have to take the square root of both sides.

    So \pm \sqrt{15} = M.
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  15. #15
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    Ah yes very good thank you.
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