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Math Help - solve each equation algebraically

  1. #1
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    solve each equation algebraically

    1/3x^2-5/6=1

    3x√ x-1 +2√ (x-1)^3=0 ......... square root symbol is over the x-1 and x-1^3

    x^3-26x^3/2=27

    2x^2(2x-1)^1/3=4x(2x-1)^4/3

    15/2^x-1=7

    3^x-1 + 3^x+1 =30

    Any help on any of thse problems would be appreciated!
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  2. #2
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    Re: solve each equation algebraically

    I can't read the first, third, fourth, fifth and sixth questions. Some brackets would be nice... LaTeX would be even nicer...

    \displaystyle \begin{align*} 3x\sqrt{x-1} + 2\sqrt{(x - 1)^3} &= 0 \\ 3x\sqrt{x - 1} + 2\sqrt{x-1}\sqrt{(x - 1)^2} &= 0 \\ 3x\sqrt{x - 1} + 2\sqrt{x - 1}|x - 1| &= 0 \\ \sqrt{x - 1}\left(3x + 2|x - 1|\right) &= 0 \\ \sqrt{x - 1} = 0 \textrm{ or } 3x + 2|x - 1| &= 0 \end{align*}

    Go from here...
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  3. #3
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    Re: solve each equation algebraically

    Hello, etran3358!

    I'll guess what the last one means . . .


    3^x-1+3^x+1=30

    3^x/3+3.3^x=30

    10/3.3^x=30

    3^x=9

    3^x=3^2

    x=2

    If my solution is hard to read and understand,
    it's because I typed it the way you did ...
    no parentheses, no spaces.


    This is what I meant . . .

    We are given: .3^{x-1} + 3^{x+1} = 30

    which equals: .(3^x)/3 + 3(3^x) = 30

    add: .(10/3)(3^x) = 30

    Multiply by 3/10: .3^x = 9

    Hence: .3^x = 3^2

    Therefore: .x = 2
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  4. #4
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    Re: solve each equation algebraically

     1/3 x^2-5/6x=1 is #1
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  5. #5
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    Re: solve each equation algebraically

    #3 is  x^3 -26x ^ 3/^ 2 = 27 it's 26 x raished to the 3/2 power
    Last edited by etran3358; August 12th 2012 at 09:42 AM.
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  6. #6
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    Re: solve each equation algebraically

     2x^2 (2x-1) ^1 / ^3 = 4x(2x-1) ^ 4 / ^3
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  7. #7
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    Re: solve each equation algebraically

     15 / 2 ^x -1 =7
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  8. #8
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    Re: solve each equation algebraically

    again i apologize if it's hard to read/understand ... i'm new to this site

    also the last problem reads as 15 ALL OVER 2^x -1 which is equal to 7

    any help would be well appreciated!
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  9. #9
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    Re: solve each equation algebraically

    i wrote these questions clearer down below...please help!!
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  10. #10
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    Re: solve each equation algebraically

    Hello, etran3358!

    You still need parentheses,
    or learn to use LaTeX more accurately.


     1/3 x^2-5/6x=1 is #1.

    Can you see that this looks like: . \frac{1}{3x^2} -= \frac{5}{6x} \:=\:1\,?

    But it might be: . \frac{1}{3}x^2 - \frac{5}{6}x \:=\:1

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  11. #11
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    Re: solve each equation algebraically

    Hello, etran3358!

    2x^2 (2x-1) ^{\frac{1}{3}} \:=\: 4x(2x-1) ^{\frac{4}{3}}

    We have: . 2x^2(2x-1)^{\frac{1}{3}} - 4x(2x-1)^{\frac{4}{3}} \:=\:0

    Factor: . . 2x\cdot(2x-1)^{\frac{1}{3}}\big[x - 2(2x-1)\big] \:=\:0

    . . . . . . . . 2x\cdot(2x-1)^{\frac{1}{3}}\cdot(2-3x) \:=\:0


    Therefore:

    . . 2x \:=\:0 \quad\Rightarrow\quad x \:=\:0

    . . (2x-1)^{\frac{1}{3}} \:=\:0 \quad\Rightarrow\quad 2x-1 \:=\:0 \quad\Rightarrow\quad x \:=\:\tfrac{1}{2}

    . . 2 - 3x \:=\:0 \quad\Rightarrow\quad x \:=\:\tfrac{2}{3}

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  12. #12
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    Re: solve each equation algebraically

    graded assignment ... thread closed
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