Algebra->Inequalities (Absolute Value)

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by Mission2012 » Tue May 07, 2013 1:11 pm
Y = |X+3| + |x-4|
Let Y = 7

7 = |X+3| + |x-4|
now testing borderlines,
test 1 -
X > 4, say X = 5
7 = 8 + 1 (false)

X < 4, say X = 3

7 = 6+1 (trues)

X = 2, 7 = 5+2 (true)

Not taking extreme value for x < 4 , say x = -100

7 = 97 + 104 = 201 (not true) hence there is point after which x < 4 doesnt hold true

Hence A is not sufficient

Second borderline, x = -3

for X < - 3, say x = - 4

7 = 1 + 8 (not true)

X > - 3

X = -2

7 = 1 + 6 (true)
X = -1

7 = 2 + 5 (ture)

X = 0,
7 = 3 + 4 (ture)

taking extreme value for x > - 3 say +100
7 = 103 + 96 = 199 (not true)

Hence B alone is not sufficient

Combine A and B, -3 < X < 4 (for x = -2, -1, 0, 1, 2 ,3 ) all confirm the equation
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by GMATGuruNY » Tue May 07, 2013 4:49 pm
Mario_87 wrote: if y = |x+3| + |4-x|, does y=7?

1. x < 4
2. x > -3
|a-b| = the DISTANCE between a and b.
|a+b| = |a-(-b)| = the DISTANCE between a and -b.

Thus:
|x+3| = the distance between x and -3.
|4-x| = the distance between 4 and x.
y = the SUM of these two distances.

Question stem rephrased:
Is the sum of the two distances equal to 7?

The distance between -3 and 4 is 7.

Thus, if x is BETWEEN these two endpoints, then the sum of the two distances will be EQUAL TO 7:
-3 <--- |x+3| ---> x <---|4-x|---> 4.
Here, |x+3| + |4-x| = the distance between -3 and 4 = 7.

By extension, if x is BEYOND either endpoint -- if x is to the left of -3 or to the right of 4 -- then the sum of the two distances will be GREATER THAN 7.

Statement 1: x < 4
If x=2, then x is between -3 and 4.
If x=-10, then x is to the left of -3.
INSUFFICIENT.

Statement 2: x > -3
If x=2, then x is between -3 and 4.
If x=10, then x is to the right of 4.
INSUFFICIENT.

Statements combined:
-3 < x < 4.
SUFFICIENT.

The correct answer is C.
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