How do I solve this complex fraction

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Oct 14th, 2015

Thank you for the opportunity to help you with your question!

You have to work your way from the bottom fraction to the top fraction :)

The answer is:

Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 14th, 2015

Because this is an easy question, we only have 20 minutes to answer. I am now going to type up the "how" of how this is done, because while the problem is easy, it's very difficulty to format in only 20 minutes :)

Oct 14th, 2015

How can I view how you got to that answer?

Oct 14th, 2015

First and foremost, this is just a series of multiplying a common denominator, combining terms, and then simplifying. 

The very first fraction to simplify is 1-2/a. We need 1 to have a common denominator before we can subtract it.

(a/a)*1 - 2/a

a/a - 2/a = a-2/a. 

Now we have:

2 + 1/(a-2/a)

Anything divided by a fraction can be multiplied by the inverse of the fraction:

1/ (a-2)/a = 1 * a/(a-2) = a/(a-2)

So now we have:

2 + a/(a-2)

We need a common denominator:

(a-2/a-2)*2 + a/(a-2) = 2a-4/(a-2) + a/a-2. Combine the numerator now that the denominator is the same:

2a-4+a/(a-2) = 3a-4/(a-2). 

So, let's state what we have now:

a + a/(2 + 1/(3a-4/a-2). 

1/(3a-4/a-2) = a-2/3a-4

2 + (a-2)/(3a-4). We need a common denominator again:

(3a-4)/(3a-4) * 2 + (a-2)/(3a-4) = 6a - 8/(3a-4) + a-2/(3a-4) = (6a - 8 + a - 2) / (3a-4)


(7a - 10) / (3a-4)

a + a/[(7a-10)/(3a-4)]

Again, anything divided by a fraction can be multiplied by its inverse:

a + a * (3a-4)/(7a-10) = a + a(3a-4)/(7a-10)

Multiply a to get the same denominator:

a*(7a-10)/(7a-10) = a(7a-10)/(7a-10) + a(3a-4)/(7a-10)


[a(7a-10) + a(3a-4) ] / (7a-10)


7a^2 - 10a + 3a^2 - 4a = 10a^2 - 14a = 2a(5a - 7)/(7a - 10)

And there you have it!

Oct 14th, 2015

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