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

I got 0,2 in both statements. Reading your responses below, I would need to substitute to get A as the answer. However, if I input( x>=3/4 ) such as x=1 in statement 1, I get 1+3=4(RHS)4(1)-3=1(LHS), any reason?

First of all, x=0 is not a solution of |x+3|=4x−3. It has only one solution x=2.

Next, the question asks whether x>0 and from (1) we get that \(x\geq{\frac{3}{4}}\), so x must be greater than 0. So, the statement answers the question. \(x\geq{\frac{3}{4}}\) does NOT mean that any value greater than or equal to 3/4 will satisfy the equation, it means that the value(s) that satisfy it must be greater than or equal to 3/4.

Hope it's clear.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution

Is x > 0?
(1) |x + 3| < 4
(2) |x - 3| < 4

In case of inequality questions, it is important to note that conditions are sufficient if the range of the question includes the range of the conditions.
There is 1 variable (x) in the original condition. In order to match the number of variables and the number of equations, we need 1 equation. Since the condition 1) and 2) each has 1 equation, there is high chance that D is the answer.
In case of the condition 1), we can obtain -4<x+3<4. This yields -7<x<1. Since the range of the question does not include the range of the condition, the condition 1) is not sufficient.
In case of the condition 2), we can obtain -4<x-3<4. This also yields, -1<x<7. Since the range of the question does not include the range of the condition, the condition 2) is not sufficient.
Using both the condition 1) and 2), since -1<x<1, the range of the question does not include the range of the conditions. Therefore, the conditions are not sufficient and the correct answer is E.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

Hi Bunuel!

Can you please tell me more/guide me towards more information about squaring both sides of an equation with absolute value? I was putting up multiple scenario equations but your solution clearly is faster. When can I square an absolute value equation? What do I need to think about?

Best wishes

When there are absolute values on both sides you can safely square. When there is an absolute value on one side and non-absolute value on another it becomes trickier - you should plug back the solutions to make sure that they are valid. You can search for similar questions here: search.php?search_id=tag&tag_id=37

Hope it helps.

Hi Bunuel,

I had a small query for this question. It will also be great if you can clarify the concept along with it.

for option B, |x-3|=|2x-3|, based on my understanding, this is the same as saying the "distance of x from 3 is same as the distance of x from 3/2". If this is true, then the value of x should fall somewhere between 3 and 1.5, i.e. 2.25 and hence should be greater than one.

Please let me know where I am wrong here

Thanks,
Dev

You would have been right if it were \(|x-3|=|x-3/2|\) but it's \(|x-3|=2*|x-3/2|\). So, the distance from x to 3 is TWICE the distance from x to 3/2.

Dear Bunuel,

Would like to ask question on the part highlighted in red in your post, specifically on determining x - 3 is negative. Since the range is \(\frac{3}{2}\leq{x}\leq{3}\), then if i plug in 3/2 in x - 3, it will result in negative number, however if i plug in 3 in x - 3, it will result in 0 or non-negative, and therefore x - 3 can be negative or non-negative. In your post x - 3 must be negative, how do we eliminate the non-negative one ? Or do we simply ignore non-negative ?

Yes, \({x}\leq{3}\), gives \(x - 3 \leq 0\) but \(|x-3|=-(x-3)\) is still true. That's because |a| = -a when a <= 0. Meaning that if a is negative or 0, then |a| = -a. For example, if say a = -1, then |a| = -a = -(-1) = 1 and if say a = 0, then |a| = -a = -0 = 0.

Does this make sense?

(1) |x + 3| = 4x – 3

RHS must be non-negative....therefore

\(4x-3\geq{0}\)............\(4x\geq{3}\)........\(x\geq{3/4}\)........Then the solution is positive

Sufficient

Dear GMATGuruNY

Is it possible to do so in statement 2 as we did in statement 1, although any side will give that x > 0? if not is it coincidence in this problem ?

Thanks in advance

Because each side of the equation in Statement 2 is an absolute value, the expressions within the absolute values (x-3 and 2x-3) do NOT have to be nonnegative.
One solution for Statement 2 is x=0, with the result that x-3 = -3 and 2x-3 = -3.

Thanks GMATGuruNY.

In this type of questions asking for sign of x, does this mean that I can deal with such statement in same manner like statement 1? or do I need open modulus to be sure from the sign?

Since the question stem asks whether x>0 -- and an equation with absolute value on both sides can have both a positive and a nonpositive solution -- I recommend that you solve the equation, either by opening up the modulus or by squaring both sides.

Bunuel does it mean that in all questions based on absolute values, one solution of it would be 0?

I'm clear with Stat1 but in stat 2, I used the formula by squaring both sides of absolute figures to find the inequality range. Here is what I did.

|x−3|=|2x−3|
Square both sides: (x−3)^2=(2x−3)^2
x^2-6x+9=4x^2-12x+9
3x^2=6x
Therefore, x=2. So I chose option D. Can you help me to find what is the error? I'm not getting it. All I did was use the formula.

1. Of course 0 is NOT a solution of all absolute value questions. Why would it be?

2. You solved 3x^2 = 6x incorrectly. You cannot reduce 3x^2 = 6x by x because x can be 0 and we cannot divide by 0. By doing so you loose a root, namely x = 0.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.

Whenever you translate an absolute value to 'go both ways' like in that explanation, you need to take an additional step and double check to make sure you haven't come up with an invalid solution. You're correct that you need to consider both x + 3 > 0 and x + 3 < 0. But once you've figured out what that implies about the value of x, double check by plugging x back in. For instance, if x + 3 < 0, we ended up figuring out that x has to equal 0. But that's impossible, because 0 + 3 isn't less than 0. So, we didn't get a valid solution at all, and only the x + 3 > 0 solution worked.

1. x = 2 works

if x<0, RHS becomes negative

Sufficient

2.
Square both sides
(2x-3)^2-(x-3)^2=0

(2x-3-x+3)(3x-6)=0

x = 0 or x=2

Not sufficient

A