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Math Help - Solving Rational Expressions

  1. #1
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    Solving Rational Expressions

    Hi, I am having trouble understanding how to answer questions like this:


    <br />
-2=<br />
\frac {1}{5}y-1<br />

    or


    <br />
\frac {-5}{2y}= \frac {1}{4}<br />
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  2. #2
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    1. Get rid of the -1 by adding 1 on bothe sides, the get rid of the x 1/5 by x5 on both sides
    2. If a/b = c/d then ad=bc (called cross multiplying) then you have a simpler equation to solve
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  3. #3
    Member rowe's Avatar
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    Quote Originally Posted by henryW View Post
    Hi, I am having trouble understanding how to answer questions like this:


    <br />
-2=<br />
\frac {1}{5}y-1<br />

    or


    <br />
\frac {-5}{2y}= \frac {1}{4}<br />
    I assume you want to solve for y?

    -2=\frac {1}{5}y-1

    -1=\frac {1}{5}y

    5\cdot(-1)=5\cdot\frac {1}{5}y

    -5=\frac {5 \cdot 1}{5}y

    -5= y

    For the second,

    \frac {-5}{2y}= \frac {1}{4}

    \frac{1}{-5}\cdot\frac {-5}{2y}= \frac {1}{4}\cdot\frac{1}{-5}

    \frac{1}{2y}= \frac {1}{-20}

    Take the reciprocal of both sides:

    2y = -20

    y = -10


    That's one way of doing it, you might not be comfortable with using reciprocals, though.




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  4. #4
    MHF Contributor

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    In general, the simplest way to handle equations with "rational fractions" is to multiply both sides by the least common denominator of all the fractions, thus elminating the fractions.

    For example, to solve 3- x/5= 1/3, multiply both sides of the equation by 3(5)= 15 to get 3(15)- (x/5)(15)= (1/3)(15) or 45- 3x= 5. Then -3x= 5- 45= -40 and x= 40/3.

    Warning, if your fractions have the "unknown", x, in a denominator, you must check any solutions in the original equation to see if any make a denominator 0. If so, that solution must be thrown out.
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  5. #5
    Super Member
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    Debsta's advice is good.
    You want to simplify the equations.
    The critical part is that whatever you do to the left hand side you must also do the equivalent to the right hand side.

    .
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