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

Question

Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

Bunuel · Accepted Answer

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.

(2)  |x+1|=2x-1  --> the same here -->  2x-1\geq{0}  -->  x\geq{\frac{1}{2}} , hence  x>0 . Sufficient.

So you see that you don't even need to find exact value(s) of  x  to answer the question.

Answer: D.

Hope it helps.

fameatop · Answer

Statement 1 If |x + 3| >= 0 -----> x+3 = 4x-3 ----->3x=6----->x=2 If |x + 3| < 0 -----> -x-3 = 4x-3 ----->5x=0----->x=0 x=0 is not possible because we assumed that x+3<0 or x<-3. Thus the only possible value of x is 2. Sufficient Statement 2 If |x + 1| >= 0 -----> x+1 = 2x-1 ----->x=2----->x=2 If |x + 1| < 0 -----> -x-1 = 2x-1 ----->3x=0----->x=0 x=0 is not possible because we assumed that x+1<0 or x<-1. Thus the only possible value of x is 2. Sufficient Answer D

amitjash · Answer

Whenever i see this type of question i start to solve for X. Now in this case, from st-1 i got X = 2 or 0 From st-2 also the same answer and i ticked E. Can someone explain how i should start to think for this type of questions?? however i forgot to put back tha values and check.

saxenashobhit · Answer

I did same and got tricked. Can someone explain why we are getting x=0 and it doesn't fit the equation. Are we doing anything wrong on absolute problems?

shrouded1 · Answer

When you are dealing with absolute values, and you solve it in 2 cases assuming first that expression >= 0 and then that expression < 0, you must verify the solution you get to make sure you don't violate the extra assumption you made in the beginning Eg.When you assume x+3>0, it means you assume x>-3. You get an answer 2, which is valid as no assumptions are violated When you assume x+3<0, it means you assume x<-3. You get an answer 0, which is invalid given your assumption You do not need to necessarily "plug the solution back in", it is sufficient to check against the assumption you made while deciding the sign of the absolute value

Bunuel · Answer

Solving for x is not the best way to deal with this problem. But if you do this way you should know that checking validity of the solutions is very important for absolute value questions. When you solve these kind of questions with check point method (discussed at: some-inequalities-questions-93760.html?hilit=substitute%20method), you test validity of solution on the stage of obtaining these values (by checking whether the value is in the range you are testing at the moment) and if the value obtained IS in the range you are testing at the moment, you don't need to substitute it afterwards to check , you've already done the checking and if the value obtained IS NOT in the range you are testing at the moment you also don't need to substitute it afterwards to check , you've already done the checking. When you just expand the absolute value once with negative sign and once with positive sign then you should check whether obtained solution(s) satisfy equation be substituting solution(s) back to the equation. Check 3-steps approach in Wlaker's post on Absolute Value for more on this: math-absolute-value-modulus-86462.html Hope it helps.

jlgdr · Answer

Is it because you have 2x - 1 and therefore 2x>=1, x>=1/2? What if you had 2x + 1? Will it still work? Don't thimk so right? We would need to analyze both scenarios Please advice Cheers! J

Bunuel · Answer

We could still solve  |x + 1| = 2x + 1  with this approach to get x. LHS is an absolute value, which is always non-negative, so RHS must also be non-negative -->  2x+1\geq{0}  -->  x\geq{-\frac{1}{2}} . For,  x\geq{-\frac{1}{2}} ,  x+1>0 , hence  |x + 1| = x + 1  -->  x + 1 = 2x + 1  -->  x=0 .

But if it were  |x - 1| = 2x + 1 , then getting  x\geq{-\frac{1}{2}}  from LHS, wouldn't be useful to determine whether  |x - 1| = x - 1  or  |x - 1| = -(x - 1)  because  x-1  could be positive as well as negative for different x'es which are more than or equal to -1/2. Therefore we should use usual method of solving for  |x - 1| = 2x + 1 , not RHS/LHS.

So, this method is not universal and can be applied only when non-modulus part defines x in such way that helps to get the sign of absolute value expression on the other side.

Does this make sense?

dabaobao · Answer

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

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