Absolute Value ( Question Bank)

[Nugs]

Is |x| < 1 ?

(1) |x + 1| = 2|x - 1|

(2) |x - 3| > 0

How would we do this question. I find Absolute questions very confusing...Can some one please explain the basics to solving this question...

[RonPurewal]

Is |x| < 1 ?

(1) |x + 1| = 2|x - 1|

(2) |x - 3| > 0

How would we do this question. I find Absolute questions very confusing...Can some one please explain the basics to solving this question...

statement (1)
for EQUATIONS involving absolute value, like this one, the key realization is that the absolute value of a quantity can signify either the quantity itself or the opposite of the quantity. therefore, if you try each of the sign combinations (pos/neg) of the absolute values in the problem, you'll be guaranteed to find all of the solutions.
(note: in what follows, "+" means leaving the expression within the absolute value bars alone; "-" means reversing the sign of that expression)
in this equation, there are ostensibly four sign combinations, +/+, +/-, -/+, -/-, but it's only necessary to try two of them:
** first, either +/+ or -/-, in which both or neither of the absolute value expressions are flipped. may as well go with +/+ (i.e., leaving both of the absolute value expressions alone while removing the bars): x + 1 = 2(x - 1), or x = 3. plugging this back into the original equation shows that it works.
** second, either +/- or -/+, in which one of the absolute value expressions is flipped. let's go (at random) with flipping the first one: -x - 1 = 2(x - 1), or x = 1/3. plugging this into the original equation also shows that it works.

therefore, statement 1 means that x = 3 or x = 1/3.

--

statement (2)

two ways to interpret absolute value inequalities like this one:
** memorize the template of the solution (preferred for efficiency's sake): you should just know that |expression| > a means "either expression > a or expression < -a".
** conceptualize absolute value as distance: in this case, |x - 3| means the distance between x and 3. therefore, this statement means that the distance between x and 3 is greater than 0 (in either direction).
either of these interpretations means that x < 3 or x > 3, or, equivalently, x is not equal to 3.

statement 1 is insufficient, because 1/3 gives a "yes" and 3 gives a "no". statement 2 is also insufficient, because every number except 3 is possible. taken together, though, the two statements are sufficient because they yield a unique value, 1/3, for x.
notice that there's no reason even to figure out whether 1/3 gives "yes" or "no" at this point; it's one value, meaning that it is guaranteed to be sufficient no matter what the answer.

answer = c

[DCE]

I think we made a typo here

for statement two the final conclusion should be:

x < - 3 or x > 3

Therefore the correct answer should be E

@Ron - Thanks for the amazing explaination above - you are the man :D

Thanks
DCE[/b]

[DCE]

Please ignore my previous post - I goofed up in calculation.

[RonPurewal]

Please ignore my previous post - I goofed up in calculation.

ok, i'll pretend it doesn't exist.
what previous post?
heh heh

[jayant.apte]

Hi Ron,

If this were the case: |x + 1| = 2|x - 1| we would test for +,+ and -,+

But what if this were the case: |x+3| - |4-x| = |8+x|, which sign combination would you test for? And how would you derive it? I am guessing that we would end up with quite a few sign combinations which would take a long time. What would be a better way to solve absolute value equations with three or more absolute value expressions?

Thanks!

[Ben Ku]

Hi Ron,

If this were the case: |x + 1| = 2|x - 1| we would test for +,+ and -,+

But what if this were the case: |x+3| - |4-x| = |8+x|, which sign combination would you test for? And how would you derive it? I am guessing that we would end up with quite a few sign combinations which would take a long time. What would be a better way to solve absolute value equations with three or more absolute value expressions?

Thanks!

The short answer is probably not. Because you have three absolute values you're adding and subtracting, very few of them overlap.

(a) +, +, +
(b) +, +, -
(c) +, -, +
(d) -, +, +
(e) +, -, -
(f) -, +, -
(g) -, -, +
(h) -, -, -

You'll notice that in your particular problem, (b) and (g) would be the same; (c) and (f) would be the same; (d) and (e) would be the same.

So really, you have (a), (b), (c), (d), and (h).

Hope that helps.