Fraction problems - Urgent

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August 1st 2006, 10:42 PM
Chuck_3000

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

$\frac{64a^3+b^3}{16a^2-b^2} / \frac{16a^2b^2-4ab^3+b^4}{4a^2-ab+12a-3b}$
August 2nd 2006, 12:49 AM
CaptainBlack

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:

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

Now cancell the common terms where possible:

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

Multiply the two denominators, and do any remaining cancellations:

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

RonL
August 2nd 2006, 01:40 AM
Chuck_3000

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

do you have any idea for the next one?
August 2nd 2006, 02:14 AM
CaptainBlack

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:

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

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:

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

Now cancel where we can:

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

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:

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

Finally cancelling:

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

RonL