RR wrote:
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.
Quote:
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)!
Quote:
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
:(
