i.ahmed111 Wrote:Is there an elegant way to quickly attack inequality problems like these, or is a more conceptual number picking approach preferred, since you are dealing with two variables and more than 1 inequality?
Thanks!
personally, i find number picking to be MORE "elegant", but that's not the issue.
you're not really dealing with two variables here. y isn't really a "variable" of its own; it's just an annoying name for the quantity |x + 3| + |4 - x|. so really, there's only one variable in this problem.
i.e., the problem is really just
If x is an integer, then is |x + 3| + |4 - x| = 7 ?(we don't even need the stipulation that y is an integer, since that's guaranteed if x is an integer)
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the algebraic method requires that you
recognize where these quantities change signs.
BACKGROUND FACT:
if "QUANTITY" is POSITIVE OR ZERO, then |QUANTITY| = QUANTITY. i.e., strip the absolute-value bars, but leave the quantity alone.
if "QUANTITY" is NEGATIVE, then |QUANTITY| = -(QUANTITY). i.e., strip the bars and flip ALL the signs.therefore:
consider |x + 3|
if (x + 3) is POSITIVE OR ZERO, then this will just be x + 3.
if (x + 3) is NEGATIVE, then this will be -x - 3.
so this is:
x + 3,
for all x that are -3 or greater-x - 3,
for all x that are less than -3consider |4 - x|
if (4 - x) is POSITIVE OR ZERO, then this will just be 4 - x.
if (4 - x) is NEGATIVE, then this will be -4 + x.
so this is:
4 - x,
for all x that are 4 or less-4 + x,
for all x that are greater than 4so if you have the two statements together, then it's just (x + 3) + (4 - x), or 7.
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of course, it's
MUCH easier just to plug in numbers, as you did, on this one.
not even close.