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Math Help - equation with fractions problem

  1. #1
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    Smile equation with fractions problem

    1
    - x - 5 =
    2

    1
    - x + 3 *(I can't write out the equation horizontally without the layout screwing up!)
    4

    I have no idea what to do with equations and fractions set out like this. Do i need to cancel them out of the equation before i make the first step to solving the unknown?

    thanks
    J
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  2. #2
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    e^(i*pi)'s Avatar
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    You need to make it clear what you mean.

    For example is it: \dfrac{11}{x-5} = \dfrac{24}{x+3}
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  3. #3
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    the layout of the equation is the fraction (1/2) x - 5 = (1/4) x + 3

    *brackets being the fraction
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  4. #4
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    Quote Originally Posted by johnnerz View Post
    the layout of the equation is the fraction (1/2) x - 5 = (1/4) x + 3

    *brackets being the fraction
    multiply both sides by 4 to clear the fractions ... then solve for x.
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  5. #5
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    e^(i*pi)'s Avatar
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    That's a better layout.

    I would go about this problem by multiplying each term by the lowest common denominator of 2 and 4 to clear the fraction.
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  6. #6
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    So is the first step (1/2) x -5 *4 = (1/4) x + 3 * 4 ?
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  7. #7
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    Quote Originally Posted by johnnerz View Post
    So is the first step (1/2) x -5 *4 = (1/4) x + 3 * 4 ?
    4\left(\dfrac{1}{2}x - 5\right) = 4\left(\dfrac{1}{4}x+3\right)
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  8. #8
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    Quote Originally Posted by johnnerz View Post
    So is the first step (1/2) x -5 *4 = (1/4) x + 3 * 4 ?
    \frac{1}{2} x - 5 = \frac{1}{4} x + 3

    4(\frac{1}{2} x - 5) = 4(\frac{1}{4} x + 3)

    Can you continue? Literally multiply each term by 4.
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  9. #9
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    x= 8 ?
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  10. #10
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    Quote Originally Posted by johnnerz View Post
    1
    - x - 5 =
    2

    1
    - x + 3 *(I can't write out the equation horizontally without the layout screwing up!)
    4

    I have no idea what to do with equations and fractions set out like this. Do i need to cancel them out of the equation before i make the first step to solving the unknown?

    thanks
    J
    Fractions are not as bad as the publicity they get...

    \displaystyle\left(\frac{1}{2}\right)x-5=\left(\frac{1}{4}\right)x+3\Rightarrow\left(\fra  c{1}{4}\right)x+\left(\frac{1}{4}\right)x-5=\left(\frac{1}{4}\right)x+3

    Therefore

    \displaystyle\left(\frac{1}{4}\right)x-5=3\Rightarrow\left(\frac{1}{4}\right)x=5+3

    A quarter of what number is 8 ?
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  11. #11
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    Quote Originally Posted by johnnerz View Post
    x= 8 ?
    4\left(\dfrac{1}{2}x - 5\right) = 4\left(\dfrac{1}{4}x + 3\right)

    2x - 20 = x + 12

    subtract x from both sides; add 20 to both sides

    x = 32


    you shouldn't have to ask if a solution is correct ... always check solutions by substituting them into the original equation to see if it makes it a true statement.
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