If x and y are nonzero integers [EIVQB:First Negative Power]

by **karim.morgan.nehdi** Sun Aug 09, 2009 5:08 pm

Hi -

I'm struggling to figure out a step that is taken in the answer explanation for a problem called "First Negative Power" in the EIV Question Bank. The problem is:

If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ?

(1) x = 2y
(2) x + y > 0

My question is about the third step in the following simplication of the question:

Can anyone please explain how you get from [(1/x)+(1/y)]^-1 on the left-hand side of the second equation to [(x+y)/(xy)]^-1 in the third equation?

Thanks.

by **venkat_yj** Mon Aug 10, 2009 7:54 pm

There isn't any step missing between equation 2 and 3 for the left-hand side.

by **karim.morgan.nehdi** Mon Aug 10, 2009 8:49 pm

venkat_yj Wrote:There isn't any step missing between equation 2 and 3 for the left-hand side.

There's probably not anything missing, I just don't get it. How do you manipulate (1/x)+(1/y) into (x+y)/(xy)?

by **venkat_yj** Mon Aug 10, 2009 9:08 pm

a/b+c/d = ((axd)+(bxc))/(bxd)........Algebra

by **john.cappuccitti** Tue Aug 11, 2009 2:59 pm

Cross multiply to get a common denominator so that you can add the two terms together. For simplicity, I'm just going to ignore the exponent and demonstrate the manipulation:

1/x + 1/y

(in order to add these together, we need a common denominator)

= (y/y)(1/x) + (x/x) (1/y)

(remember, multiplying by x/x or y/y is the same as multiplying by 1, so it does not change the original equation)

= y/xy + x/xy

(now just add them together because they have the same denominator)

= (x+y)/xy

by **Ben Ku** Tue Aug 18, 2009 12:54 pm

Thanks John, that's good work. Karim, let me know if you still have problems about this question.

by **melissae.brown** Wed Sep 09, 2009 10:40 am

in this question, can you get rid of the -1 exponent from the beginning?

by **Ben Ku** Fri Oct 16, 2009 3:00 am

melissae.brown:

Re: If x and y are nonzero integers [EIVQB:First Negative Power]
in this question, can you get rid of the -1 exponent from the beginning?

I think what you're asking is if we can just do this:
Is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ?
Is (x^-1 + y^-1) > [(x^-1)(y^-1)]?

Well, we know that x^-1 = 1/x. So I will re-state the question, using a and b instead of the complicated expressions.
Is a^-1 > b^-1?
Is 1/a > 1/b?

Now we cannot really simplify any further because we don't know the sign (positive or negative) of a or b. Therefore we cannot rephrase the original question to
Is a > b?

Hope that makes sense.