The Reason Behind Absolute Value Questions on the GMAT

VeritasKarishma ,
I still don't understand WHY |x| = -x when x<0.
To my understanding |x| is the distance of x from zero and distances are not negative. So, then why?

What is l-7l, it is the distance of -7, on the number line, from zero.
No let's follow the formula lxl = -(x) iff x<0,
Here x=-7, cool.
so answer is - (-7), i.e. 7.

So, the absolute value (read distance from zero on the number line) of -7 is 7

Hope this helps. Karmesh

You are ignoring the second part of the statement: "when x < 0"
When x itself is NEGATIVE, -x is POSITIVE. Of course, distance is positive. Hence when x itself is positive, |x| is x but when x itself is negative, you need to make it positive by making it -x.

i dont understand why negative is assigned in case 3, when x>4, could you please explain?­

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When x > 4, then 4 - x < 0, thus |4 - x| = -(4 - x).

Here is a complete solution:

If x and y are integers and y = |x + 3| + |4 - x|, does y equal 7?

\(y=|x+3|+|4-x|\) two check points: \(x=-3\) and \(x=4\) (check point: the value of \(x\) when expression in || equals to zero), hence three ranges to consider:

A. \(x<{-3}\) --> \(y=| x + 3| +|4-x| =-x-3+4-x=-2x+1\), which means that when \(x\) is in the range {-infinity,-3} the value of \(y\) is defined by \(x\) (we would have multiple choices of \(y\) depending on \(x\) from the given range);

B. \(-3\leq{x}\leq{4}\) --> \(y=|x+3|+|4-x|=x+3+4-x=7\), which means that when \(x\) is in the range {-3,4} the value of \(y\) is \(7\) (value of y does not depend on value of \(x\), when \(x\) is from the given range);

C. \(x>{4}\) --> \(y=|x+3|+|4-x|=x+3-4+x=2x-1\), which means that when \(x\) is in the range {4, +infinity} the value of \(y\) is defined by \(x\) (we would have multiple choices of \(y\) depending on \(x\) from the given range).

Hence we can definitely conclude that \(y=7\) if \(x\) is in the range {-3,4}


(1) \(x<4\) --> not sufficient (\(x<4\) but we don't know if it's \(\geq{-3}\));
(2) \(x>-3\) --> not sufficient (\(x>-3\) but we don't know if it's \(\leq{4}\));

(1)+(2) \(-3<x<4\) exactly the range we needed, so \(y=7\). Sufficient.

Answer: C.

OR: looking at \(y=|x+3|+|4-x|\) you can notice that \(y=7\) (\(y\) doesn't depend on the value of \(x\)) when \(x+3\) and \(4-x\) are both positive, in this case \(x-es\) cancel out each other and we would have \(y=|x+3|+|4-x|=x+3+4-x=7\). Both \(x+3\) and \(4-x\) are positive in the range \(-3<{x}<4\) (\(x+3>0\) --> \(x>-3\) and \(4-x>0\) --> \(x<4\)).

Hope it's clear.­

Understood sir. Thank you for taking the time to explain.
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