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Thread: Basic algebra

  1. #1
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    Basic algebra

    1) 1=12+n

    2) 3=-5-10x

    3) 4=5/3n

    4) 8^-1/y +3^-1=-3

    5) 10c=-3-3c

    6) 6=-10-5m

    7) 11=y/2^-1 -7

    8) 10/11=3-8y+8

    9) 5=60n^-1 -1

    10) 4/8=10n/6
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by jerryramos View Post
    1) 1=12+n
    Subtract 12 from both sides to get: $\displaystyle n=1-12=11$

    Try the other ones, and then tell me what ones you need help on...
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jerryramos View Post
    1) 1=12+n

    2) 3=-5-10x

    3) 4=5/3n

    4) 8^-1/y +3^-1=-3

    5) 10c=-3-3c

    6) 6=-10-5m

    7) 11=y/2^-1 -7

    8) 10/11=3-8y+8

    9) 5=60n^-1 -1

    10) 4/8=10n/6
    2) 3=-5-10x

    5 + 3 = 5 + -5-10x

    8 = -10x

    $\displaystyle \frac{8}{-10} = \frac{-10x}{-10}$

    $\displaystyle x = -\frac{8}{10} = -\frac{4}{5}$

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jerryramos View Post
    3) 4=5/3n
    $\displaystyle 4 = \frac{5}{3n}$

    $\displaystyle 3n \cdot 4 = 3n \cdot \frac{5}{3n}$

    $\displaystyle 12n = 5$

    $\displaystyle \frac{12n}{12} = \frac{5}{12}$

    $\displaystyle n = \frac{5}{12}$

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jerryramos View Post
    4) 8^-1/y +3^-1=-3
    Note that [tex]a^{-1} = \frac{1}{a}[tex] ($\displaystyle a \neq 0$.)

    So
    $\displaystyle \frac{8^{-1}}{y} +3^{-1}=-3$

    $\displaystyle 8^{-1} \cdot \frac{1}{y} +3^{-1}=-3$

    $\displaystyle \frac{1}{8} \cdot \frac{1}{y} + \frac{1}{3} = -3$

    Let's get rid of these pesky fractions, shall we? The least common multiple of 3 and 8 is 24, so multiply both sides of the equation by 24:

    $\displaystyle 24 \left ( \frac{1}{8} \cdot \frac{1}{y} + \frac{1}{3} \right ) = 24 \cdot (-3)$

    $\displaystyle 3 \cdot \frac{1}{y} + 8 = -72$

    $\displaystyle 3 \cdot \frac{1}{y} + 8 - 8= -72 - 8$

    $\displaystyle 3 \cdot \frac{1}{y} = -80$

    $\displaystyle 3 \cdot \frac{1}{y} \cdot y = -80 \cdot y$

    $\displaystyle 3 = -80y$

    $\displaystyle \frac{3}{-80} = \frac{-80y}{80}$

    $\displaystyle y = -\frac{3}{80}$

    -Dan
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