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Simplify: $$\frac{2t-t^2}{t+2} \cdot \left(\frac{5t}{t-2} - \frac{2t}{t-2} \right)$$

  1. I first subtracted the parenthesis because the denominator is equal. I then got:

$$\frac{2t-t^2}{t+2} \cdot \frac{3t}{t-2}$$

  1. Then I was lost. I tried multiplying by $t+2$ and $t-2$ on either sides. I tried multiplying $3t$ with $-1$. To make $t+2$ on both sides. Again I didn't get the answer. I have so many different calculations that I'm lost.

None of these seem correct. Or I multiplied incorrect but I doubt that. Is there a trick for doing these? I'm wasting lots of time on just one simplification. Just when I think I'm progressing I'm stuck again.

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  • $\begingroup$ actualy it is t-2 instead of t/2 in the paranthesis. $\endgroup$
    – user160137
    Commented Jul 6, 2014 at 18:34
  • $\begingroup$ Is that correct now? $\endgroup$
    – amWhy
    Commented Jul 6, 2014 at 18:39
  • $\begingroup$ Yes, thank you. $\endgroup$
    – user160137
    Commented Jul 6, 2014 at 18:39

3 Answers 3

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$$\frac{2t-t^2}{t+2} \cdot (\frac{5t}{t-2} - \frac{2t}{t-2} )=\frac{2t-t^2}{t+2} \cdot\frac{3t}{t-2} =\frac{-t(-2+t)}{t+2}\cdot\frac{3t}{t-2}=\frac{-3t^2}{t+2}$$

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  • $\begingroup$ I didn´t see 2t-t 2² could be expanded further. (-2+t) can be elimated with (t-2). And now it makes sense. Thank you! $\endgroup$
    – user160137
    Commented Jul 6, 2014 at 18:45
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$$\begin{align} \frac{2t-t^2}{t+2}\cdot\left(\frac{5t}{t-2}-\frac{2t}{t-2}\right) &= \frac{2t-t^2}{t+2}\cdot \frac{3t}{t-2} \\ &= \frac{3t\left(2t-t^2\right)}{(t+2)(t-2)} \\ &= \frac{6t^2-3t^3}{(t+2)(t-2)} \\ &= \frac{-3t^2(t-2)}{(t+2)(t-2)} \\ &= -\frac{3t^2}{t+2} \\ \end{align}$$

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  • $\begingroup$ That's what I did! 6t²-3t³. I really need to see those expanding steps myself. I tried it but didn´t saw that it could be expanded. Also, thank you! $\endgroup$
    – user160137
    Commented Jul 6, 2014 at 18:51
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$$ \frac{3t}{t/2} = \frac{t\cdot 3}{t\cdot\frac 1 2} = \frac{3}{1/2} = 3\cdot2. $$

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