Bunuel wrote:
If y = |x – 1| and y = 3x + 3, then x must be between which of the following values?
(A) 2 and 3
(B) 1 and 2
(C) 0 and 1
(D) –1 and 0
(E) –2 and –1
MANHATTAN GMAT OFFICIAL SOLUTION:We should set the two equations for y equal and algebraically solve |x – 1| = 3x + 3 for x.
This requires two solutions: one for the case that x – 1 is positive, the other for the case that x –1 is negative:
x – 1 is positive: (x – 1) = 3x + 3 --> x = -2.
INCORRECT: x – 1 = (–2) – 1 = –3
Therefore, the original assumption that x – 1 is positive does not hold.
x – 1 is negative: –(x – 1) = 3x + 3 --> x = -1/2.
CORRECT: Therefore, the original assumption that x – 1 is negatives holds.
Thus, there is only one solution for x and y, which is y = 3/2 and x = -1/2.
The correct answer is D.So I solved for y initially to find the distance. One answer was -3 and the second for 3/2. Since distance from 1 cannot be negative, only 3/2 is valid.
But why then can it not be between 2 and 3 as well? Is that because it x as 5/2 needs to satisfy y=3x+3? That is the only reason I could think of. The possible answer I solved for was -1/2 and 5/2 (distance of 3/2 from 1 but only one of them is correct).