Simplifying Expression 2

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July 9th 2009, 08:31 AM
Sooda

Simplifying Expression 2

Made new thread because I got new expression to solve. I have tried to get answer by myself, but ended with different result.

Expression:
$(\frac{2x}{x+2}+\frac{2x}{6-3x}+\frac{8x}{x^2-4})\div\frac{4x^2+16x}{x-2}$

Answer:
$\frac{1}{3(x+4)}$

I have done this:
$(\frac{2x*-3(x-2)+2x(x+2)+8x*-3}{-3(x+2)(x-2)})$

$(\frac{2x*-3(x-2)+2x(x+2)+8x*-3 *(x-2)}{-3(x+2)(x-2)*(4x^2+16x)})$

Here (x-2) should cancel each other out which leaves:
$(\frac{2x*-3(x-2)+2x(x+2)+8x*-3}{-3(x+2)(4x^2+16x)})=\frac{24x^2-44x-4x^3+48}{48x^2-192x-12x^4+48x^3}$

Pretty nice, when I made x=1 and solved my gigantic simplification it gave same result as text book simplification: 1/15

My creation blows, where I have done mistakes? How this needs to be done?
July 9th 2009, 09:23 AM
alexmahone

$(\frac{2x}{x+2}+\frac{2x}{6-3x}+\frac{8x}{x^2-4})\div\frac{4x^2+16x}{x-2}$

= $2x\ (\frac{1}{x+2}+\frac{1}{6-3x}+\frac{4}{x^2-4}) * \frac{x-2}{4x(x+4)}$

= $(\frac{1}{x+2}-\frac{1}{3(x-2)}+\frac{4}{x^2-4}) * \frac{x-2}{2(x+4)}$

= $\frac{3(x-2)-(x+2)+4*3}{3(x^2-4)} * \frac{x-2}{2(x+4)}$

= $\frac{3x-6-x-2+12}{3(x-2)(x+2)} * \frac{x-2}{2(x+4)}$

= $\frac{2x+4}{3(x-2)(x+2)} * \frac{x-2}{2(x+4)}$

= $\frac{2(x+2)}{3(x-2)(x+2)} * \frac{x-2}{2(x+4)}$

= $\frac{1}{3(x+4)}$
July 9th 2009, 10:35 AM
Sooda

Thank you very much, now I can learn from your simplification. Bringing 2x in front of expression is new for me.

EDIT:
When you do this:
$\frac{1}{6-3x}=-\frac{1}{3(x-2)}$

Will it only affect denominator and its argument places? When you change term sign then arguments in denominator get swapped? Will numerator stay unchanged by that move? Can I read somewhere about that operation?
July 9th 2009, 10:41 AM
allyourbass2212

Quote:

Originally Posted by Sooda

Thank you very much, now I can learn from your simplification. Bringing 2x in front of expression is new for me.

Same, I was not aware you could factor in that manner.
July 9th 2009, 02:52 PM
yeongil

Quote:

Originally Posted by Sooda

EDIT:
When you do this:
$\frac{1}{6-3x}=-\frac{1}{3(x-2)}$

Will it only affect denominator and its argument places? When you change term sign then arguments in denominator get swapped? Will numerator stay unchanged by that move? Can I read somewhere about that operation?

Two different things are going on here. First, you have this subtracting property: a - b = -(b - a). Swapping the order of subtraction requires you to put a negative outside. So what was done in the above was that
$\frac{1}{6-3x}=\frac{1}{-(3x - 6)}=\frac{1}{-3(x - 2)}$ .

Next, you have a fraction where the denominator is negative. When dealing with negatives, remember that the following fractions are equivalent:
${\color{red}-}\frac{a}{b} = \frac{{\color{red}-}a}{b} = \frac{a}{{\color{red}-}b}$ . The negative sign in our problem was in the denominator, and can be moved to in front of the fraction:
$\frac{1}{{\color{red}-}3(x - 2)}={\color{red}-}\frac{1}{3(x - 2)}$ .

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