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Math Help - Simplifying Expressions, having trouble with this one

  1. #1
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    Simplifying Expressions, having trouble with this one

    (3)(x - 7) / (x + 2) (2x - 1) multiplied by (4x^2 + 6x - 4) / (4x - 28)

    I could use step by step help on how to simplify this expression.
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  2. #2
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    The second fraction, factor out the 4 in the denominator and 2 in the numerator. Then multiply out the denominator in the first fraction and you should be able to move on from there.
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  3. #3
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    Quote Originally Posted by jenbones View Post
    (3)(x - 7) / (x + 2) (2x - 1) multiplied by (4x^2 + 6x - 4) / (4x - 28)

    I could use step by step help on how to simplify this expression.
    \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{4x^2+6x-4}{4x-8}

    just a bit of factoring in each step ...

    \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{2(2x^2+3x-2)}{4(x-2)}

    \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{2(2x-1)(x+2)}{4(x-2)}

    fully factored, can you finish?
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  4. #4
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    Quote Originally Posted by skeeter View Post
    \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{4x^2+6x-4}{4x-8}

    just a bit of factoring in each step ...

    \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{2(2x^2+3x-2)}{4(x-2)}

    \frac{3(x-7)}{(x+2)(2x-1)} \cdot \frac{2(2x-1)(x+2)}{4(x-2)}

    fully factored, can you finish?
    I am confused one what to do next ...
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  5. #5
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    Quote Originally Posted by jenbones View Post
    I am confused one what to do next ...
    The original expression is \displaystyle\frac{3(x-7)}{(x+2)(2x-1)}\ . \frac{4x^2+6x-4}{4x-28}


    \displaystyle\frac{3(x-7)}{(x+2)(2x-1)}\ . \frac{2(2x-1)(x+2)}{4(x-7)}=\frac{2x-1}{2x-1}\ . \frac{x+2}{x+2}\ . \frac{2(3)}{4}\ . \frac{x-7}{x-7}
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    The original expression is \displaystyle\frac{3(x-7)}{(x+2)(2x-1)}\ . \frac{4x^2+6x-4}{4x-28}


    \displaystyle\frac{3(x-7)}{(x+2)(2x-1)}\ . \frac{2(2x-1)(x+2)}{4(x-7)}=\frac{2x-1}{2x-1}\ . \frac{x+2}{x+2}\ . \frac{2(3)}{4}\ . \frac{x-7}{x-7}
    I cancel out (x + 2) / (x + 2) and (x - 7) / (x - 7) next right?
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  7. #7
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    Quote Originally Posted by jenbones View Post
    I cancel out (x + 2) / (x + 2) and (x - 7) / (x - 7) next right?
    Yes, but don't forget about the 3rd set!
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  8. #8
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    Quote Originally Posted by Archie Meade View Post
    Yes, but don't forget about the 3rd set!
    you mean (2x - 1) / (2x -1 ) ?
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  9. #9
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    Quote Originally Posted by jenbones View Post
    you mean (2x - 1) / (2x -1 ) ?
    Yes, the three fractions containing x, where the numerator and denominators are identical are all just the value 1,
    (though strictly speaking, we'd have to say that the function is undefined for values of x giving zero in the denominator).

    So the value of the fraction is simply 6/4=1.5.
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