x > 1 Quantity A: x / (x+1)

Cydney Seigerman, Magoosh Tutor

Hi Charles!

Happy to help :) Because we know that x > 1, we can actually cross multiply to solve this question. However, it can be a little tricky.

Since x > 1, then the denominator of A is positive while the denominator of B is negative. So, when we cross multiply, we are multiplying across an inequality by a negative number. That means we are actually reversing the sign of the inequality (even though we don't know yet if the inequality is <, >, or =). We will need to consider this reversal, though, as we solve this question.

Originally we have

x / (x+1) ? -x / (1 -x)

where ? denotes that we don't know the relationship between the two values.

Once we cross multiply, we have

x(1-x) ¿ -x(x+1)

where ¿ indicates that we've reversed the sign of the original inequality.

From here, we can multiply through to get rid of the parenthesis on both sides:

x - x^2 ¿ -x^2 - x

Let's cancel the term (-x^2) on both sides now:

x ¿ -x

Since x > 1, then x > -x. However, remember that this is the opposite of the actual relationship. So, we can conclude the A < B. And this is the same conclusion that Chris reached using the method in the explanation video :)

I hope this helps!

Jul 30, 2017 • Reply

Francisco Olivas

Thanks Cydney, very helpful. So for applying this method, this is the ETS method right?, once we reverse the inequality in the last step, to reach the final answer we can ALWAYS do what you did of going from x>-x to A < B?

Oct 11, 2017 • Reply

Sam Kinsman

Hi Francisco,

My name is Sam, and I'm happy to help you on Cydeny's behalf!

I'm not quite sure how the ETS solved this practice problem, since the PDF practice book where this came from doesn't have answer explanations.

Let's go over the last few steps in the inequality again. We have:

x ¿ -x

Notice that we haven't reversed the inequality sign yet. Now, we can plug in a value for x. We know is greater than 1, so let's plug in 2:

x ¿ -x
2 ¿ -2

Well, the left hand side is bigger. So quantity A is greater.

However, we still have to reverse the inequality sign! (That's why we have the "¿" sign). So instead of stating that Quantity A is bigger, we know Quantity B must be bigger.

I hope that helps!

Nov 8, 2017 • Reply

Ishita Singh

Hi Sam, I am a little confused with the last step- once we determine quantity A is greater, why are we reversing the inequality sign?

Jan 12, 2018 • Reply

David Recine

Hi Ishita,

I can answer that for Sam. :) We reverse the inequality sign here because of one simple rule:

If you cross multiply a negative value across an inequality, you need to reverse the sign. Quantity B includes negative x, so the sign will be reversed once cross-multiplication with Quantity A is complete.

Jan 12, 2018 • Reply

Pooja Saha

Why can this not be solved algebraically by solving for x? I did this and got x ? 0, so I assumed A was larger without plugging in any numbers.

Nov 4, 2014 • Reply

Anthony Mosca

I'm actually wondering the same exact thing as Pooja. I completely understand the "plug-in" method in the video, but when I try to algebraically simplify both sides by multiplying each numerator by the other side's denominator and simplifying through cancellation of like terms, I arrive at 2x for Quantity A and 0 for Quantity B. I figure I'm performing some illegal algebraic move, but I can't figure out what it is.

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Update to my previous comment:

I realize that, by multiplying by both quantities (A and B) by the other quantity's denominator, I was multiplying both sides by a negative number (the denominator of B). As a result, the final inequality that results is essentially flipped. This more convoluted method does concur with the "plug-in" method.

Nov 11, 2014 • Reply

Seung Gee Kim

what if x is a fraction?

Nov 30, 2016 • Reply

Sam Kinsman

Hi Seung,

If x is a fraction, Col. B will still always be greater than Col. A. Keep in mind that x has to be greater than 1. So we can have any fraction that is greater than 1 (i.e. the numerator is greater than the denominator).

I'd suggest that you test a few different fractions. You'll find that for each possible value that x could have, Col. B will still be greater! :)

Dec 7, 2016 • Reply