Ab Val/Inequality DS

[james]

If y >= 0, what is the value of x?

1) |x-3| >= y

2) |x-3| =< -y


Answer is B, but I have no idea why. Any help would be greatly appreciated. Thank you!

[RR]

Given y >= 0

i. |x-3| >= y
y is positive and |x-3| is also positive. So for any value of y, we can have a value of x such that |x-3| will be greater than y. INSUFFICIENT

ii. |x-3| =< -y
We know that y is greater than or equal to zero. Therefore
RHS : -y has to be negative or zero
LHS : |x - 3| has to be greater than or equal to zero (Since it is absolute value)
Since LHS is positive and LHS <= RHS, the RHS cannot be negative.
There RHS has to be zero ie -y = 0 ie. y = 0
Since LHS <= RHS, and the LHS cannot be less than zero, the LHS = 0
ie |x - 3| = 0 ie x = 3. SUFFICIENT

Answer is B

Now, that said, this is not how I started out doing the problem. I started out by solving with equations. When I see problems with absolute values and algebra, I am always in a confusion whether to approach it logically or through equations. Here is how I first approached the problem.
i. |x-3| >= y
Case I :
x - 3 >= y
x >= y + 3
Case II :
-(x - 3) >= y
x -3 <= -y
x <= 3 - y
Therefore (3 - y) >= x >= (y + 3). INSUFFICIENT

ii. |x-3| =< -y
Case I:
x - 3 =< -y
x =< 3 - y
Case II:
-(x - 3) =< -y
x - 3 >= y
x >= y + 3
Therefore (3 - y) >= x >= (y + 3). INSUFFICIENT

Note here that both i & ii give the same answer.

I am not sure where I went wrong. If someone could clarify, it would really help.

[RonPurewal]

this just may be the single most posted problem on this forum.

here's a sampler plate:

http://www.manhattangmat.com/forums/pos ... html#10352

http://www.manhattangmat.com/forums/if- ... t2675.html

http://www.manhattangmat.com/forums/if- ... -t895.html

search, people!

[RR]

Ron, thank you for the reply. None of the solutions outline a pure algebraic approach. While I understand that the logical method is the quicker way to do it, would appreciate if you could tell me where I went wrong with my algebraic approach.

i. |x-3| >= y
Case I :
x - 3 >= y
x >= y + 3
Case II :
-(x - 3) >= y
x -3 <= -y
x <= 3 - y
Therefore (3 - y) >= x >= (y + 3). INSUFFICIENT

ii. |x-3| =< -y
Case I:
x - 3 =< -y
x =< 3 - y
Case II:
-(x - 3) =< -y
x - 3 >= y
x >= y + 3
Therefore (3 - y) >= x >= (y + 3). INSUFFICIENT

Note here that both i & ii give the same answer.

[RonPurewal]

Ron, thank you for the reply. None of the solutions outline a pure algebraic approach. While I understand that the logical method is the quicker way to do it, would appreciate if you could tell me where I went wrong with my algebraic approach.

comment #1: it's easy to type "<" or ">". just type a normal "<" or ">" using the underline tags.
"<=" and ">=" are somewhat difficult to read - especially the former, which looks like an arrow. (and "=<" looks like a frowny face)

--

you should definitely try to incorporate this sort of casewise reasoning into your treatment of absolute-value problems; you really, really don't want to do pure algebra.
in fact, on problems like this, you HAVE to incorporate logic into your solution; there's no way to get around having to realize that -y is a negative quantity. there just isn't.

i. |x-3| >= y
Case I :
x - 3 >= y
x >= y + 3
Case II :
-(x - 3) >= y
x -3 <= -y
x <= 3 - y
Therefore (3 - y) >= x >= (y + 3). INSUFFICIENT

first of all, you can't combine these two inequalities, ever.
if you have an absolute value that's greater than some number ("|expression| > a"), that produces a disjunction: i.e., two inequalities joined by "OR", not "AND".
so this should produce the disjunction x < 3 - y OR x > y + 3.
this is the principal problem here.

analogy: take a simple inequality, such as |x| > 5.
the solution to this inequality is "x > 5 OR x < -5", not "-5 > x > 5" (which is not only incorrect, but also impossible).

also note that your sandwich inequality, (3 - y) > x > (y + 3), is impossible unless y = 0. (y is not allowed to be negative, and, if y > 0, the right-hand quantity is bigger than the left-hand quantity.) in that case you'd have 3 > x > 3, so x would have to be 3.
(as before, you can't realize this with a "pure algebraic approach".)
therefore, ironically, based on the inequality you wrote above, you should have concluded that this statement is sufficient (even though it isn't)!

ii. |x-3| =< -y
Case I:
x - 3 =< -y
x =< 3 - y
Case II:
-(x - 3) =< -y
x - 3 >= y
x >= y + 3
Therefore (3 - y) >= x >= (y + 3). INSUFFICIENT

this time your algebra is correct; inequalities of the form "|expression| < a" really DO produce sandwich inequalities.
BUT
as noted above, this sandwich inequality requires y + 3 to be less than or equal to 3 - y, which means that y must be < 0.
since y is not allowed to be negative, this means y = 0.
therefore 3 > x > 3.
therefore x = 3.
sufficient.
(same conclusion you should have reached above, based on your incorrect inequality - only this time your inequality is correct, so it's actually sufficient this time)

--

summary:
the "logic" you mention is not some tricky shortcut; it's essential to the problem.
trying to ignore it in favor of a "purely algebraic approach" is nonsensical; you just can't.
same thing with inequalities / number props problems, such as, say, xz < yz; you can't address that inequality "algebraically" without considering the different cases corresponding to positives and negatives.
you just can't.

sorry
:(

[RR]

Thank you very much Ron for the detailed explanation. Realized the folly now ! Feel like a duffer :)

[RR]

And yes, shall henceforth use < and > :)[/u]

[RonPurewal]

And yes, shall henceforth use < and > :)[/u]

glad to hear this.

now if everyone else follows your example, and if we can also manage get rid of "+ve", "-ve", "shud", and the like, i'll be a very happy forum chief.