Manipulating Fractions

Isktaine · #1 $Old$ 11-15-2008, 01:33 PM

I'm trying to do an intergration using the t-substitution so I've been reading examples and I keep loosing track on a similar fraction problem. I don't understand how
$\frac{4}{3+(10t/(1+t^2))} * \frac{2}{1+t^2} = \frac{8}{3t^2+10t+3}$

Any insight on how you can manipulate this is greatly appreciated

Cheers,
Isk

Math_Helper · #2 $Old$ 11-15-2008, 01:52 PM

o_O · #3 $Old$ 11-15-2008, 01:52 PM

${\color{red}\frac{4}{3 + \displaystyle \frac{10t}{(1+t^2)}}} \ \cdot \ {\color{blue}\frac{2}{1+t^2}}$

To multiply fractions, simply multiply numerator by numerator and denominator by denominator:

$= \frac{{\color{red}4} \cdot {\color{blue}2}}{{\color{red}\left(3 + \displaystyle \frac{10t}{(1+t^2)}\right)}{\color{blue}\left(1+t^2\right)}}$

You should know that: $a(b+c) = ab + ac$ . Here, imagine $a = {\color{blue}1+t^2}$ .

So, simplifying:
$= \frac{8}{{\color{red}3}{\color{blue}(1+t^2)} + \displaystyle \frac{{\color{red}10t}}{{\color{blue}1+t^2}} {\color{blue}\left(1+t^2\right)}}$

Notice that in the denominator, the $1 + t^2$ cancels out in the second term, so we get:

$= \frac{8}{3(1+t^2) + 10t}$

and so on and so on ..

Isktaine · #4 $Old$ 11-15-2008, 02:00 PM

Thank you soo much!