GMATinsight wrote:
The sum of the possible values of X in the equation |X + 7| + |X – 8| = 16 is:
A. 0
B. 1
C. 2
D. 3
E. None of the above
If you want to solve by removing brackets (not bad time-wise, 1:50): there are four cases to check.
Let A stand for (X + 7)
Let B stand for (X - 8).
There are four cases to consider for the expressions, just exactly as if the two absolute values were on opposite sides of the equation:
+A +B
-A +B
+A -B
-A -B
Evaluate equation with all four cases:
1. (X + 7) + (X - 8) = 16
X + 7 + X - 8 = 16
2X - 1 = 16
2X = 17
X = \(\frac{17}{2}\)
2. (-X - 7) + (X - 8) = 16
-X - 7 + X - 8 = 16. Stop.
-X + X = 0. We cannot get a value for X
3. (X + 7) + (-X + 8) = 16
X + 7 - X + 8 = 16. Stop. Not a solution, same as #2
4. (-X - 7) + (-X + 8) = 16
-X - 7 - X + 8 = 16
-2X + 1 = 16
-2X = 15
X = -\(\frac{15}{2}\)
Check each solution in original equation. When we solve this way, this step is imperative.
|\(\frac{17}{2}\) + 7| + |\(\frac{17}{2}\) - 8| = 16
|\((\frac{17 + 14)}{2}\)| + |\(\frac{(17-16)}{2}\)| = 16
|\(\frac{31}{2}\)| + |\(\frac{1}{2}\)| =
\(\frac{32}{2}\) = 16 Correct
The other solution, -\(\frac{15}{2}\), also works. (Check it.)
The sum of the two solutions?
\(\frac{17}{2}\) + (-\(\frac{15}{2})\)
= \(\frac{2}{2}\) = 1
Answer B
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