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Math Help - Solve for x

  1. #1
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    Solve for x

    How would you solve this for  x ??

    <br />
\frac{\frac{1}{x+1}-\frac{1}{4}}{x-3} = 0<br />
(fixed it! made it into an equation)

    i think you multiply the 1 from the numerator to the denominator, but im not sure if you can do that when there is a subtract sign between the two integers. help??
    Last edited by kdl00; March 3rd 2010 at 05:58 PM.
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  2. #2
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    Quote Originally Posted by kdl00 View Post
    How would you solve this for  x ??

    <br />
\frac{\frac{1}{x+1}-\frac{1}{4}}{x-3}<br />

    i think you multiply the 1 from the numerator to the denominator, but im not sure if you can do that when there is a subtract sign between the two integers. help??
    first of all, x cannot be solved for since what you have written is not an equation, but an expression.

    the expression, however, can be simplified.

    get a common denominator and combine the fractions in the numerator, then divide the resulting fraction by (x-3)
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  3. #3
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    Quote Originally Posted by skeeter View Post
    first of all, x cannot be solved for since what you have written is not an equation, but an expression.

    the expression, however, can be simplified.

    get a common denominator and combine the fractions in the numerator, then divide the resulting fraction by (x-3)
    actually, this was a limit question in calc., but i wanted to know how to get rid of the fraction on the numerator, thats why i posted it in algebra.
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  4. #4
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    is the fraction equal to 0 or are you trying to find the limit of the fraction as x \to 0 ?
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  5. #5
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    This expression is indeterminate. The only way a fraction can equal zero is if the numerator equals zero. The only way this numerator could equal zero would be if x was equal to 3, but that would make the denominator equal to zero as well.
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  6. #6
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    ok ... I saw your limit post in the calculus section.

    \lim_{x \to 3} \frac{\frac{1}{x+1} - \frac{1}{4}}{x-3}<br />

    same procedure as I described before for simplification ...

    \lim_{x \to 3} \frac{\frac{4}{4(x+1)} - \frac{1(x+1)}{4(x+1)}}{x-3}<br />

    \lim_{x \to 3} \frac{\frac{4- (x+1)}{4(x+1)}}{x-3}<br />

    \lim_{x \to 3} \frac{\frac{3-x}{4(x+1)}}{x-3}<br />

    \lim_{x \to 3} \frac{-1}{4(x+1)} = -\frac{1}{16}<br />

    next time, be straightforward about what you want.
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  7. #7
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    Quote Originally Posted by skeeter View Post
    ok ... I saw your limit post in the calculus section.

    \lim_{x \to 3} \frac{\frac{1}{x+1} - \frac{1}{4}}{x-3}<br />

    same procedure as I described before for simplification ...

    \lim_{x \to 3} \frac{\frac{4}{4(x+1)} - \frac{1(x+1)}{4(x+1)}}{x-3}<br />

    \lim_{x \to 3} \frac{\frac{4- (x+1)}{4(x+1)}}{x-3}<br />

    \lim_{x \to 3} \frac{\frac{3-x}{4(x+1)}}{x-3}<br />

    \lim_{x \to 3} \frac{-1}{4(x+1)} = -\frac{1}{16}<br />

    next time, be straightforward about what you want.
    kk got it. thanks for the help
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