How do you write a complex fraction that when simplified results in 1/x?

1 Answer
Nov 7, 2017

Follow the intuition below to get an answer, but one sample solution is #((x)/(x+3))/((x^2)/(x+3))#

Explanation:

A complex fraction involves having two fractions inside one. Keeping this in mind, all you need to do is put together expressions that give you the above.

Here's the approach I'd suggest:

Note that dividing two fractions is the same thing as multiplying by the reciprocal. So say your complex fraction is #(a/b)/(c/d)#. This would be the same thing as:

#(a/b) * (d/c)#.

This makes things slightly easier to look at.

Now, you know that the numerator of your resulting fraction must equal 1. So, the best course of action is to set #a = c# so they cancel out to make #1#.

Next, you know that the denominator of the resulting fraction must equal #x#. So, you should set #b = d* x#, so it simplifies down to #x#.

#((x)/(x+3))/((x^2)/(x+3))# is one such example. If we wrote this out as a product:

#x/(x+3) * (x+3)/x^2#

You see that the #(x+3)# terms just cancel out to give you 1. Also, the #x# and #x^2# divide out to give you an #x# in the denominator.

Hope that helped :)