Manhattan DS Is X>0?

[getso]

Is x > 0?

(1) |x + 3| = 4x -3

(2) |x + 1| = 2x -1

I always get stuck in such problems.

@Testluv:Your inputs on this problem would be of great help.

Thanks,
Shobha

[Testluv]

Absolute value is always positive or zero.

(1) |x + 3| = 4x -3

We see absolute value on the left. Thus, the left hand side must be positive or zero. Because the left hand side must equal the right hand side, this means that the right hand side is also positive or zero. Thus, 4x-3>= 0, and x >=3/4. Therefore, x is definitely positive. So, the first statement is sufficient by itself. Similar reasoning applies to the second statement.

Both statements sufficient by themselves; choose D.

[yserious]

Absolute value is always positive or zero.

(1) |x + 3| = 4x -3

We see absolute value on the left. Thus, the left hand side must be positive or zero. Because the left hand side must equal the right hand side, this means that the right hand side is also positive or zero. Thus, 4x-3>= 0, and x >=3/4. Therefore, x is definitely positive. So, the first statement is sufficient by itself. Similar reasoning applies to the second statement.

Both statements sufficient by themselves; choose D.

Hi TestLuv,

Isn't taking the absolute value as if it were not an then taking both teh positive an negative to it?
i.e (x+3) = 4x-3 Case 1?
-(x+3) =4x-3 Case 2? (though in this case the answer may not gave the correct value when plugged back into the original question..)

Should absolute values without taking the +ve and -ve values be taken into consideration always?

[Testluv]

Hi yserious


If we had this: |x| = 4, then x is either 4 units to the right or 4 units to the left of zero on the number line, and so either x = 4 or x = -4, and we definitely consider negative values.

But you would never have an equation like this: |x| = -4 because, as a measure of distance, absolute value is always non-negative. Having an equation like that is exactly the same as saying "one thing is negative four units distant from another thing". Obviously, the closest two objects can get to each other is perfectly joining each other, in which case the distance between them is zero; the distance between them cannot be negative. Thus, if we have |x| = y, we know the left hand side is positive or zero, and therefore the right hand side is positive or zero; that is, y is positive or zero.

So, we can look at statement one like this:

(1) |x + 3| = 4x -3

4x - 3 = |x + 3|

4x - 3 = (pos or zero)

4x - 3 >= 0

One tip-off that the question was looking to be approached like this is that it was asking you for the sign of x. Then, it put statements with absolute value. So, then think about the possible signs of absolute value: either positive sign or no sign.

[pink_08]

Since absolute value include +ve and -ve, Why are we not considering the -ve case

x + 3 = - ( 4x -3)

x + 3 = 3- 4x

x=0

[Testluv]

Since absolute value include +ve and -ve, Why are we not considering the -ve case

x + 3 = - ( 4x -3)

x + 3 = 3- 4x

x=0

Normally, that would be a great approach but here "x" appears on both sides, and so there is a risk that one of the cases is impermissible, that it will result in a negative absolute value. For example, if x = 0, and we sub in to the equation of statement one, we have:

|0 + 3| = 4(0) - 3

|3| = -3

and that equation is impossible because absolute value is non-negative.

So you could have employed pure algebra but then you would have had to test the validity of your numerical results against the known equation(s).

Many absolute value and inequality questions are better approached through reasoning about the relevant conepts rather than pure algebra. And many others benefit from employing both conceptual reasoning and algebra (such as here).

(Late in this thread, I employed a similar approach on a question involving an inequality of fractions : https://www.beatthegmat.com/gmat-prep-ii ... tml#208044. And here is another absolute value question where I employed a similar approach: https://www.beatthegmat.com/value-of-y-t ... tml#201008)

[pink_08]

Great stuff testlov !

Is there kind of a clear separation on what problems to approach through reasoning and what problems thro pure algebra

[Testluv]

Great stuff testlov !

Is there kind of a clear separation on what problems to approach through reasoning and what problems thro pure algebra

Well, I would say as a very very crude rule of thumb: the harder the problem, the more likely it will be attackable through reasoning.

However, the broader theme here is that you should be open to alternative approaches in general. The test-maker is always looking to reward test-takers who arrive at deductions that obviate the need for unnecessary algebra. When reviewing your work, make sure you review your correct answers too (not just your incorrect ones), and always ask yourself: "was there a quicker way I could have gotten to the right answer? Could I have picked numbers, backsolved, used commonsense, logical reasoning, etc?"

[getso]

Hi Testluv,

Thanks for the wonderful explanation.

Regards,
Shobha

[linkinpark]

Hi Testluv,
I agree with whatever you explained for this problem, however I've one question

will I be correct if I squared of both side of equations and solve them to check x >0?
e.g.
|x+1| = 2x - 1
(x+1)^2 = (2x-1)^2 and then just solve, applying this to both stmt1 and 2.

[Testluv]

Hi Testluv,
I agree with whatever you explained for this problem, however I've one question

will I be correct if I squared of both side of equations and solve them to check x >0?
e.g.
|x+1| = 2x - 1
(x+1)^2 = (2x-1)^2 and then just solve, applying this to both stmt1 and 2.

Yes, you can because |x| = sq root of x^2.

But how would you go about applying this to the first statement? How ugly would all of that math be and how long do you think it would take?

I think you have to compare the time it would take to do all of that to how much time you think it would take solving it the way I did in my very first post on this thread.

[amittilak]

Since absolute value include +ve and -ve, Why are we not considering the -ve case

x + 3 = - ( 4x -3)

x + 3 = 3- 4x

x=0

When I approached this problem, I did exactly what you did above and got values of x =0 or x=2. However, once you get the values, you need to plug it back into the original equation to make sure that both of them satisfy the equation. This is especially true for Absolute value eqn's, with variables on both sides. x=0 does not satisfy the conditions (l3l cannot be equal to -3) and hence we need to discard it. x=2>0 is suff.
For S2, x=0 or x=2, which is same as S1. Hence, suff.
This process takes more time than the conceptual approach suggested by the other poster.
However, if you are like me and find it easier to solve eqns quickly, you can use the approach I like to use.

[maihuna]

Is x > 0?

(1) |x + 3| = 4x -3

(2) |x + 1| = 2x -1

Just doing for my own practice:

1. for x< -3 x+3 < 0 so -(x+3) = 4x - 3 so 5x = 0 or x=0, but since x<-3, x=0 is not applicable.
for x>=3, x+3 > 0 so x+3 = 4x-3 => 3x = 6 or x = 2. So X>0

2. for x<-1 -(x+1) = 2x-1 or 3x = 0 => x=0 not applicable.
for x>=-1, x+1 = 2x-1 = > x=2

so D

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But tesluv.. hz can absolute values always be positive or zero. I did not understand tat..

[Testluv]

But tesluv.. hz can absolute values always be positive or zero. I did not understand tat..

Absolute value measures the distance of something from zero (on the number line). Distance is always positive (you can't say that something is negative meters away from something else, for example). So, take "-5". How far away from zero is it?...5 units. So, the absolute value of "-5" is 5.

So, if you have:

|x|=y,

you know that:

x = pos or zero

So, y is positive or zero.