they're being tricky little #$@^&*'s here.

but first, i must address this:

guest612 wrote:

I flip the inequalities sign when I divide by the equation by -1. Thus, (3+x) is greater than or equal to y.

no no no! no!

and by the way, no!

the expression in question is an ABSOLUTE VALUE, |x - 3|. two things about this expression:

(1) you

ABSOLUTELY CANNOT remove the absolute value bars and replace all the minus signs with plus signs (i.e., produce x + 3). this will always, always, ALWAYS be wrong.

if you don't see what's so wrong about this, then compare |10 - 3| and 10 + 3; that ought to get the idea across. remember, any statement that's false for numbers is also false for variables, because variables stand for numbers.

(2) what it IS equivalent to is either

(x - 3), if x - 3 is nonnegative (i.e., if x

> 3), or

(-x + 3), the opposite of the expression within the bars, if x - 3 is negative (i.e., if x < 3).

if you don't know the value of x, there is

no way to remove the absolute value bars, so don't try.

instead...

think about the

number properties associated with absolute values. in particular, the most important property of absolute values - which, fortunately, is also one of their simplest and most easily understood - is this:

absolute values can't be negative.
this observation alone is actually enough to solve this problem. here's how:

(1): all this tells you is that |x - 3|, an absolute value, is at least some positive number. this isn't much of a restriction, as it could be anything greater than that number, so there's no way you'll get a unique value for x. insufficient.

(2): (-y) is 0 or less, and

the absolute value must be at or below this value.

combined with the basic observation that appears in boldface above, this tells you that

y must be 0, and that the absolute value itself must also be 0. (think about the other possibilities for (-

y) if this doesn't make sense to you right away). therefore, |x - 3| = 0, so x must be 3. sufficient!