Absolutely less than zero (2)

[Vikram]

Is x > 0?

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

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

The described solution goes on to picking numbers and substituting them in the question to find the value of x..

Alternatively, would this be a correct way to solve this?

a) 4x - 3 >=0
4x >= 3
x >= 3/4 Hence Sufficient.

b) 2x - 1 >=0
2x >= 1
x >= 1/2 Hence sufficient.

This is the same as the correct answer, D. Now my question is, is this a valid way to solve this. Am I missing something. Would there be any question where following this approach would put me in a spot of bother?

Vikram.

[jster]

Can you only do this assumption that the left side is greater than zero when there is a variable on that side?

Say the question was |x+3| = 11 .... then would you have to do the following

x>0
x+3 = 11
x = 8

OR

x<0
-(x+3) = 11
-x -3 = 11
-x = 14
x = -14

So insufficient?

[JK]

I don't think you are suppose to use the left side without absolute values to say when it is greater than zero...instead do the following

1) |x+3| = 4x-3
x+3 >= 0 when x= > -3
x+3 < 0 when x < -3

for x >= -3 when x+3 >= 0
x+3 = 4x-3
x = 2

for x < -3 when x+3 < 0
-(x+3) = 4x+3
x = 0, this statement cannot be true since we are using x < -3
SUFFICIENT

2) |x+1| = 2x-1
x+1 >= 0 when x >= -1
x+1 < 0 when x < -1

for x+1 >= 0
x+1 = 2x-1
x = 2

for x+1 < 0
-(x+1) = 2x -1
x = 0, this cannot be true since we are using x < -1
SUFFICIENT

D is your answer

[RonPurewal]

Is x > 0?

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

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

The described solution goes on to picking numbers and substituting them in the question to find the value of x..

Alternatively, would this be a correct way to solve this?

a) 4x - 3 >=0
4x >= 3
x >= 3/4 Hence Sufficient.

b) 2x - 1 >=0
2x >= 1
x >= 1/2 Hence sufficient.

This is the same as the correct answer, D. Now my question is, is this a valid way to solve this. Am I missing something. Would there be any question where following this approach would put me in a spot of bother?

Vikram.

so yeah, you get lucky here. what you're doing is narrowing down the possibilities, in what i must admit is a rather clever way, but please do not think for even 1 second that you've actually solved the equations.
specifically, it's not at all true that x > 3/4 is the actual solution of the first equation, nor is x > 1/2 the actual solution of the second. however, you have narrowed the possibilities: it's impossible for x to solve the first equation unless x > 3/4, and it's impossible for x to solve the second equation unless x > 1/2.
those conditions happen to settle the issue in this case, but this sort of reasoning is certainly not generalizable. see below.

--

to actually SOLVE these equations, you just set the absolute value equal to (1) the expression within the bars, and (2) the opposite of the expression within the bars. then solve, CHECK, and keep the solutions that work.

(1)
first, x + 3 = 4x - 3
6 = 3x
x = 2
check: this works.

then, -x - 3 = 4x - 3
0 = 5x
x = 0
this doesn't work (it gives 3 = -3).

so, x = 2 is the only solution.
if the problem had said "is x > 3/2?", you would mistakenly think it was insufficient, when it's in fact sufficient. in fact, because this equation yields only one solution, it's automatically sufficient, regardless of the actual question (it's impossible for one number to give two answers to a question prompt).

(2)
first, x + 1 = 2x - 1
2 = x
check: this works

then, -x - 1 = 2x - 1
3x = 0
this doesn't work

so, x = 2 is the only solution.
also sufficient, for the same reasons as mentioned above.
and, same pitfalls as mentioned above re: your solution.

--

are you saying that the actual given solutions use number picking? are you serious? you can't pick numbers for these sorts of things; they have definite solutions. it's almost impossible to hit upon the correct solutions at random.

[malikrulzz]

Hi Ron,
That was grt.

What would have been the case if there would have been Mod on both the sides?
Say for ex
|x| = |2| even then can we treat it as x = +/ - 2

[malikrulzz]

Question is asking Is x > 0?

however from both the statements we are getting

x=0 which is not the case
x=2 which statisfis the eq.

IMO E. what you suggest?

[Guest]

Correct Ans should be D.

Each is sufficient.
From 1)
If x+3 >0;x>-3 Then from 1) x=2 and this does not violate the condition

If x+3<0;x<-3;Then from 1) x=0...The assumption is false hence x=2>0

From 2) If x+1>0;x>-1

Now x+1=2x-1

Or x=2;Hence true...the other Gives if x+1<0;x<-1;Then x+1=-2x+1=>x=0 which is False...Hence x=2>0...Sufficient

[esledge]

What would have been the case if there would have been Mod on both the sides?
Say for ex
|x| = |2| even then can we treat it as x = +/ - 2

Yes, malikrulzz, that would be correct. Actually, your example can be simplified before you start dealing about the absolute value of x, as you know that |2| = 2. So |x| = 2 means x = +2 or -2.

More generally, for each set of absolute value bars, you want to think of two cases for what's inside: positive and negative.

Take this for example: |2x - 3| = |4x + 1|.

There are 4 cases
(A) positive inside each set of bars: (2x - 3) = (4x + 1)
(B) positive inside left set, negative inside right: (2x - 3) = (-4x - 1)
(C) negative inside left set, positive inside right: (3 - 2x) = (4x + 1)
(D) negative inside each set of bars: (3 - 2x) = (-4x - 1)

If you solve each of the 4 equations individually, you get
(A) (2x - 3) = (4x + 1) --> -4 = 2x --> x = -2
(B) (2x - 3) = (-4x - 1) --> 6x = 2 --> x = 1/3
(C) (3 - 2x) = (4x + 1) --> 2 = 6x --> x = 1/3
(D) (3 - 2x) = (-4x - 1) --> 2x = -4 --> x = -2

It is typical to get duplicate solutions and occasionally invalid solutions, so you must plug x = -2 and x = 1/3 back into the original to verify.
x = -2: |2(-2) - 3| = |4(-2) + 1| --> |-4 - 3| = |-8 + 1| -->|-7|=|-7| CORRECT.
x = 1/3: |2(1/3) - 3| = |4(1/3) + 1| --> |2/3 - 3| = |4/3 + 1| = |-7/3| = |7/3| CORRECT.

Therefore, x = -2 or 1/3.

Unfortunately, you can't take a shortcut. You must set up and solve all 4 cases, as you won't know which one(s) give(s) valid answers.

Question is asking Is x > 0?

however from both the statements we are getting

x=0 which is not the case
x=2 which statisfis the eq.

IMO E. what you suggest?

Reread Ron's post and notice that when he checked, 0 was found to be an invalid solution to the original equations (in the statements). Thus, it gets thrown out and forgotten. I think this happened because there was only one set of absolute value bars, so you were effectively "stuck with" the sign of the expression on the right. The only valid solution is 2.

[malikrulzz]

Thanks Emily

OK if we put x = 0
we get |3| = -3 which is is incorrect, right?

[JonathanSchneider]

Well, you'd only put in 0 for x if 0 were one of the answers that you got. Emily has described a process by which we can solve for the possible values of the variable in the absolute value bars. After solving, we must check to make sure that these values actually work. However, you don't want to start plugging in random values. 0 was not listed as a possible value for x, so don't plug that one in. Make sense?