Hi, there. I'm happy to help a bit with this.

QUESTION #1Yes, it's true that, as long as a=/0 and b=/0, then |a + b| =< |a| + |b|. That's an actual math rule. First of all, notice it's not a "less than" sign, but a "less than or equal to" sign --- that's very important. This idea is closely related to something Geometry known as the Triangle Inequality. The basic logic is ---if a and b are both positive, or both negative, then the two sides will be the same, but if a and b have opposite sign, they will get subtracted on the left side, while their absolute values will be added on the right side. Example |(-3) + 5| = |2| = 2, but |(-3)| + |5| = 3 + 5 = 8. Does that make sense?

Caveat: this is always true, but not necessarily all that useful in problem-solving. If you have a DS question asking explicitly about comparing |a + b| to |a| + |b|, then this formula would be relevant, but in a general absolute value question, you are much better off enumerating cases, as you say you do anyway.

QUESTION #2Yes, using a number line to mark +/- cases is the standard and most widely used method of approaching quadratic inequalities. It is a 100% valid and legitimate approach. If this method makes sense to you, there is no disadvantage to using it every single time.

QUESTION #3OK, a DS question.

Prompt:

Is a + a^-1 > 2?Here's an algebraic approach. The big idea is --- when you have an x + 1/x, that is, an x to the +1 and an x to the -1, those are exponents that differ by two --- that means, in hidden form, you are dealing with a quadratic. That's what I will bring out in the solution below.

Subtract 2 from both sides:

a - 2 + 1/a > 0

Factor out an 1/a

(1/a)*(a^2 - 2a + 1) > 0 we have the square of a binomial!

(1/a)*(a - 1)^2 > 0

So, of these two factors, (1/a) is positive when a > 0 and negative when a < 0. By contrast, (a - 1)^2 is positive for all real number a EXCEPT when a = 1 --- that makes the expression equal to zero.

Statement #1:

a > 0The expression will be positive for all values of a except at a = 0, in which case the left side equals zero, and the inequality is false. So, a continuous infinity of possible values of a would make the prompt question true, and only one would make it false, but that's enough to say we can't answer the prompt definitively. Statement #1 by itself is

insufficient.

Statement #2:

a <1The good news is --- the dreaded value of a = 1 is avoided, so the (a - 1)^2 is always positive. BUT, now a can be less than zero, and if a is negative, then 1/a is also negative, and that would make the inequality false. Again, we don't have a definitive answer. Statement #2 by itself is

insufficient.

Statements #1 & #2 combined:

Now, we know 0 < a < 1. We avoid the dreaded a = 1 value, so (a - 1)^2 is always positive, and because a has to be greater than zero, we know that 1/a is also always positive. Therefore, the product of those two terms is always positive, and the inequality is always true. Combined, the statements are

sufficient.

I realize this is probably a very different approach from the one you took to solve this DS. If you'd like to post your work, I can help you figure out where you went wrong and what mistakes to avoid in the future.

Here's another related GMAT DS question, for more practice.

http://gmat.magoosh.com/questions/999The question at that link, when you submit your answer, will be followed by a page with a complete video explanation.

Does everything I've said this make sense? Please let me know if you have any questions on what I've said, or if I can help you in any other way.

Mike

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Mike McGarry

Magoosh Test Prep