general algerbra question

[kabtq9s]

this is a question from an old exam that I am trying to understand

I don't think you need the passage but I included it anyways along with the explanation of the answer

I specifically don't understand the 2nd step of the algebra where they have they Qc/Qc in the numerator.

http://mysowar.wordpress.com/files/2009/06/refrig-1.jpg
http://mysowar.wordpress.com/files/2009/06/refrig-2.jpg

thanks in advance

---

Members don't see this ad.

 

[sleepy425]

Here's what they did (or what they tried to do, but they have a typo):

n=Qc/(Qh-Qc)

They are simply dividing the numerator and the denominator by Qc (which you can do, because it's like multiplying by (1/Qc)/(1/Qc) which equals to 1:

n=(Qc/Qc)/[(Qh-Qc)/Qc]=1/(Qh/Qc-1)

Now, they have written n=(Qc/Qc)/(1-Qh/Qc). This is an ERROR on their part. As you can see, the expression they wrote is what you'd get if you multiplied what I wrote by (-1). In other words, that's what you'd get if the original equation were

n=Qc/(Qc-Qh)

Anyway, then everything follows the same way, you plug in the idea equation and you get:

n=1/(Th/Tc-1)

Multiply by Tc/Tc:

n=Tc/(Th-Tc)

which is choice A

So all they did in the step you asked about was that they divided the top and bottom by Qc so that you'd get Qh/Qc in the denominator, but they accidentally switched the two terms in the denominator. Hope that helps