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# Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1

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Manager
Joined: 24 Oct 2016
Posts: 72
GMAT 1: 670 Q46 V36
Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1  [#permalink]

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02 Aug 2018, 10:47
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Difficulty:

55% (hard)

Question Stats:

53% (01:37) correct 47% (01:49) wrong based on 30 sessions

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Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

_________________
Manager
Joined: 24 Oct 2016
Posts: 72
GMAT 1: 670 Q46 V36
Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1  [#permalink]

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02 Aug 2018, 10:50
1
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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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1  [#permalink]

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03 Aug 2018, 00:05
2
dabaobao wrote:
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)

This question has MUCH easier solution.

Is x > 0?

(1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) $$|x + 1| = 2x – 1$$. The same here. LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$2x-1\geq{0}$$ --> $$x\geq{\frac{1}{2}}$$, hence $$x>0$$. Sufficient.

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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x + 1| = 2x – 1 &nbs [#permalink] 03 Aug 2018, 00:05
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