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Math Help - Fraction problems - Urgent

  1. #1
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    Fraction problems - Urgent

    \frac{4x^2-1}{x^2-9y^2} * \frac{2x^2+6xy-x-3y}{4x^2-4x+1} * \frac{1}{2x+1}

    and also

    \frac{64a^3+b^3}{16a^2-b^2} / \frac{16a^2b^2-4ab^3+b^4}{4a^2-ab+12a-3b}
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Chuck_3000
    \frac{4x^2-1}{x^2-9y^2} \times \frac{2x^2+6xy-x-3y}{4x^2-4x+1} \times \frac{1}{2x+1}
    first expand all the terms which are the difference of two squares, and factorise of the denominator of the middle term:

    <br />
\frac{4x^2-1}{x^2-9y^2} \times \frac{2x^2+6xy-x-3y}{4x^2-4x+1} \times \frac{1}{2x+1} <br /> <br />
=\frac{(2x-1)(2x+1)}{(x-3y)(x+3y)} \times \frac{(x+3y)(2x-1)}{4x^2-4x+1} \times \frac{1}{2x+1}<br />

    Now cancell the common terms where possible:

    <br />
\frac{4x^2-1}{x^2-9y^2} \times \frac{2x^2+6xy-x-3y}{4x^2-4x+1} \times \frac{1}{2x+1} <br /> <br />
=\frac{(2x-1)}{(x-3y)} \times \frac{(2x-1)}{4x^2-4x+1}<br />

    Multiply the two denominators, and do any remaining cancellations:

    <br />
\frac{4x^2-1}{x^2-9y^2} \times \frac{2x^2+6xy-x-3y}{4x^2-4x+1} \times \frac{1}{2x+1} <br /> <br />
=\frac{(4x^2-4x+1)}{(x-3y)(4x^2-4x+1)}=\frac{1}{x-3y}<br />

    RonL
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  3. #3
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    thanks for that i got 1 over 3y-x, so im glad i got close

    do you have any idea for the next one?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Chuck_3000

    \frac{64a^3+b^3}{16a^2-b^2} / \frac{16a^2b^2-4ab^3+b^4}{4a^2-ab+12a-3b}
    First get rid of the division and replace it with a multiplication:

    <br />
\frac{64a^3+b^3}{16a^2-b^2} / \frac{16a^2b^2-4ab^3+b^4}{4a^2-ab+12a-3b}= <br /> <br />
\frac{64a^3+b^3}{16a^2-b^2} \times \frac{4a^2-ab+12a-3b}{16a^2b^2-4ab^3+b^4}<br />

    Now on the right hand side of the equals sign expand the difference of two
    squares in the numerator of the first term and factorise the denominator of
    the second term:

    <br />
\frac{64a^3+b^3}{16a^2-b^2} / \frac{16a^2b^2-4ab^3+b^4}{4a^2-ab+12a-3b}= <br /> <br />
\frac{64a^3+b^3}{(4a-b)(4a+b)} \times \frac{(4a-b)(a+3)}{16a^2b^2-4ab^3+b^4}<br />

    Now cancel where we can:

    <br />
\frac{64a^3+b^3}{16a^2-b^2} / \frac{16a^2b^2-4ab^3+b^4}{4a^2-ab+12a-3b}= <br /> <br />
\frac{64a^3+b^3}{(4a+b)} \times \frac{(a+3)}{16a^2b^2-4ab^3+b^4}<br />

    Now a little trial and error shows that 4a+b is a factor of 64a^3+b^3, so doing the
    long division, and factoring out the factor of b^2 in the numerator of the last term gives:

    <br />
\frac{64a^3+b^3}{16a^2-b^2} / \frac{16a^2b^2-4ab^3+b^4}{4a^2-ab+12a-3b}= <br /> <br />
\frac{(4a+b)(16a^2-4ab+b^2)}{(4a+b)} \times \frac{(a+3)}{b^2(16a^2-4ab+b^2)}<br />

    Finally cancelling:

    <br />
\frac{64a^3+b^3}{16a^2-b^2} / \frac{16a^2b^2-4ab^3+b^4}{4a^2-ab+12a-3b}= <br /> <br />
\frac{a+3}{b^2}<br />

    RonL
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