Hi, i was wondering if i simplify first or later on. If later, what step?

(x-5/2x)/(x^2/4x^2)

thank you for any help!

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- Apr 14th 2008, 11:14 AM #1

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- Apr 15th 2008, 12:52 AM #7
$\displaystyle \frac{~~~~\frac{x-5}{2x}~~~~}{\frac{x^2}{4x^2}}$

The very first thing you should see is that the x^2 cancel out in the lower fraction.

$\displaystyle \frac{~~~~\frac{x-5}{2x}~~~~}{\frac{x^2}{4x^2}}~~~~=~~~~\frac{~~~~\f rac{x-5}{2x}~~~~}{\frac{1}{4}}$

Now, when you divide by a fraction, that is the same as multiplying by it's reciprocal. For example when you divide by 1/4, you are actually multiplying by 4/1.

$\displaystyle \frac{~~~~\frac{x-5}{2x}~~~~}{\frac{1}{4}}~~~~=~~~~\frac{x-5}{2x}*\frac{4}{1}$

Now Multiply the 4 through the numerator, and since anything times 1 equals 1, we can just ignore the 1

$\displaystyle \frac{x-5}{2x}*\frac{4}{1}~~~~=~~~~\frac{4(x-5)}{2x}$

And since we are multiplying by 4, and dividing by 2, 4/2 = 2 so we can simplify this:

$\displaystyle \frac{4(x-5)}{2x}~~~~=~~~~\frac{2(x-5)}{x}$

Now you can distribute the 2 if you like, or you can leave it factored like it is, that will depend on your teacher's preference. But when you distribute, you multiply the coefficient (the 2) by each element in the term (the x-5) so we get 2(x-5) = 2x-2*5 which is 2x-10

$\displaystyle \frac{2(x-5)}{x}~~~~=~~~~\frac{2x-10}{x}$

So there is your answer. You can choose whichever you think your instructor will consider "simpler"