# dividing fractions

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• Sep 12th 2007, 12:00 PM
blame_canada100
dividing fractions
this is a fraction (x-2)^2 over 4x^2-16 divided by 21x over 3x+6 how do you divide it and make it into its lowest terms?
• Sep 12th 2007, 12:03 PM
Jhevon
Quote:

Originally Posted by blame_canada100
this is a fraction (x-2)^2 over 4x^2-16 divided by 21x over 3x+6 how do you divide it and make it into its lowest terms?

do you mean $\displaystyle \frac {\frac {(x - 2)^2}{4x^2 - 16}}{\frac {21x}{3x + 6}}$ ?
• Sep 12th 2007, 12:28 PM
blame_canada100
i mean (x-2)^2 over 4x^2-16 (thats one fraction) divided by 21x over 3x+6(thats another fraction)
• Sep 12th 2007, 12:39 PM
Jhevon
Quote:

Originally Posted by blame_canada100
i mean (x-2)^2 over 4x^2-16 (thats one fraction) divided by 21x over 3x+6(thats another fraction)

umm, yes, that's what i have

$\displaystyle \frac {\frac {(x - 2)^2}{4x^2 - 16}}{\frac {21x}{3x + 6}} = \frac {(x - 2)^2}{4x^2 - 16} \div \frac {21x}{3x + 6}$

$\displaystyle = \frac {(x - 2)^2}{4x^2 - 16} \cdot \frac {3x + 6}{21x}$

$\displaystyle = \frac {(x - 2)^2}{4 \left( x^2 - 4 \right)} \cdot \frac {3(x + 2)}{21x}$

$\displaystyle = \frac {(x - 2)^2}{4(x - 2)(x + 2)} \cdot \frac {3(x + 2)}{21x}$

Hopefully the rest is obvious to you. if not, it's ok, say so