dabaobao wrote:

Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

(1) SUFFICIENT: Here, we are told that |x + 3| = 4x – 3. When dealing with equations containing variables and absolute values, we generally need to consider the possibility that there may be two values for the unknown that could make the equation work.

We can find the two possible solutions by setting the expression inside the absolute value equal to ± the expression on the right side of the equation. In this case, our two solutions are:

x + 3 = +(4x – 3) and x + 3 = –(4x – 3)

x + 3 = 4x – 3 x + 3 = –4x + 3

6 = 3x 5x = 0

2 = x x = 0

However, you need to be careful. When there is a variable outside the absolute value, both solutions are not always valid. We need to plug both x = 2 and x = 0 back into the original equation and test them.

|(2) + 3| = 4(2) – 3 and |(0) + 3| = 4(0) – 3

|5| = 8 – 3 |3| = –3

5 = 5 3 = –3

Therefore, x = 0 is not a valid solution and we know that 2 is the only possible solution. x is definitely positive.

(2) SUFFICIENT: Here, again, we must consider the two possible solutions for the absolute value expression. They are:

x + 1 = +(2x – 1) and x + 1 = –(2x – 1)

x + 1 = 2x – 1 x + 1 = –2x + 1

2 = x 3x = 0

x = 0

Once again, we need to verify that both solutions are valid. We need to plug x = 2 and x = 0 into the original equation:

|(2) + 1| = 2(2) – 1 and |(0) + 1| = 2(0) – 1

|3| = 4 – 1 |1| = –1

3 = 3 1 = –1

One again, 2 is the only valid solution and we can determine that x is positive.

The correct answer is D. (OE:

Manhattan Prep)

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