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

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Is x > 0? (1) |x + 3| = 4x 3 (2) |x + 1| = 2x 1 Can [#permalink] New post 03 Sep 2010, 13:20
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Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

Can someone explain this to me
[Reveal] Spoiler: OA

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Re: Absolute value DS [#permalink] New post 03 Sep 2010, 13:29
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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.
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Re: Absolute value DS [#permalink] New post 03 Sep 2010, 13:43
thanks that helped your explanation is more simpler and fundamental!

the below explanation i did not get especially the last part where we need to substitute and check

(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.
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Re: Absolute value DS [#permalink] New post 03 Sep 2010, 15:15
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rxs0005 wrote:
thanks that helped your explanation is more simpler and fundamental!

the below explanation i did not get especially the last part where we need to substitute and check

(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.


This issue is discussed here: some-inequalities-questions-93760.html?hilit=substitute%20method
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Re: Absolute value DS [#permalink] New post 07 Sep 2010, 22:33
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.
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Re: Absolute value DS [#permalink] New post 19 Sep 2010, 23:42
rxs0005 wrote:
thanks that helped your explanation is more simpler and fundamental!

the below explanation i did not get especially the last part where we need to substitute and check

(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.


This appears every so often on absolute value questions. It relies on you not testing the possible solution. You must always verify the solutions you get in your first step. I've made this mistake several times :)
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Re: Absolute value DS [#permalink] New post 20 Sep 2010, 18:18
Bunuel wrote:
(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


great way to look at it. thanks for the tip
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Re: Absolute value DS [#permalink] New post 20 Sep 2010, 18:44
amitjash wrote:
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.

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?
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Re: Absolute value DS [#permalink] New post 20 Sep 2010, 19:10
saxenashobhit wrote:
amitjash wrote:
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.

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?


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
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Re: Absolute value DS [#permalink] New post 20 Sep 2010, 20:54
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saxenashobhit wrote:
amitjash wrote:
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.

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?


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.
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Re: Absolute value DS [#permalink] New post 09 Oct 2013, 06:28
Bunuel wrote:
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.


For those more inclined to solving equations, here is a different approach:
(1) |x+3|=4x-3

If x+3>0 then:
a) x > -3 AND
b) x+3 = 4x-3 => 3x=6 => x=2.
Since x=2 does NOT contradict x>-3, x=2 is a solution and x is positive

However if x+3<0 then:
a) x<-3 AND
b) x+3=-4x+3 => x=0. BUT x must be less than -3, so x=0 is NOT a solution.

Therefore only the case where x+3>0 can be possible being x=2 the only possible solution, therefore always positive.


(2) |x+1|=2x-1 Works exactly the same:

If x+1>0 then:
a) x > -1 AND
b) x+1 = 2x-1 => x = 2
Since x=2 does NOT contradict x>-1, x=2 is a solution and x is positive

However if x+1<0 then:
a) x<-1 AND
b) x+1 = -2x+1 => x = 0. BUT x must be less than -1, so x=0 is NOT a solution.

Therefore only the case where x+1>0 can be possible being x=2 the only possible solution, therefore always positive.

In BOTH cases the only possible value for x is always positive. Therefore D is the answer.
Re: Absolute value DS   [#permalink] 09 Oct 2013, 06:28
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