gmatdriller wrote:
For example, we know that the expression in an absolute sign is always >= 0
so, from (1) we can say 4x - 3 >= 0
and x >= 4/3. Also, x > 0. Sufficient
By the same token, x >= 1/2 and thus > 0
Also sufficient.
That's the fast way to do the problem, and it's perfectly correct.
[email protected] wrote:ans is ddd
from 1: |x + 3| = 4x - 3
if x > 0, than x+3 = 4x -3 by solving this we get 6= 3x. thus x =2
if x< 0, than x +3 = -4x + 3, by solving this we get x =0 . but if we put x=0 in the question than we see that |x + 3| is not equal to 4x - 3. thus x is not equal to 0
Your analysis of these cases isn't quite right. |x+3| is equal to x+3 when *what's inside the absolute value* is positive. So |x+3| = x+3 when x > -3, not when x > 0. You then find, solving, the perfectly good solution x = 2. In the second case, |x+3| = -x-3 when what's inside the absolute value is negative. That is, |x+3| = -x-3 when x < -3 (and not when x < 0; |x+3| is still equal to x+3 if you plug in x = -1, say). So in the second case, when x < -3, we'd find, solving, that x = 0, but this solution doesn't agree with our assumption that x < -3, so cannot be a valid solution. You don't need to plug the value back into the equation.
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