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

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Re: [tricky one] Is x > 0? [#permalink]  04 Jun 2013, 07:42
Expert's post
|x+3|=4x-3. Check point is at x=-3 (check point, is the value of x for which the value of an expression in modulus equals to zero).

When x\leq{-3}, then x+3<0, thus |x+3|=-(x+3). So, in this case we have -(x+3)=4x-3 --> x=0 --> discard this solution since we consider the range when x\leq{-3}.

When x>{-3}, then x+3>0, thus |x+3|=x+3. So, in this case we have x+3=4x-3 --> x=2 --> since x=2>-3, then this solution is valid.

So, we have that |x+3|=4x-3 has only one root: x=2.

Hope it's clear.
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Re: [tricky one] Is x > 0? [#permalink]  04 Jun 2013, 08:10
Bunuel wrote:
|x+3|=4x-3. Check point is at x=-3 (check point, is the value of x for which the value of an expression in modulus equals to zero).

When x\leq{-3}, then x+3<0, thus |x+3|=-(x+3). So, in this case we have -(x+3)=4x-3 --> x=0 --> discard this solution since we consider the range when x\leq{-3}.

When x>{-3}, then x+3>0, thus |x+3|=x+3. So, in this case we have x+3=4x-3 --> x=2 --> since x=2>-3, then this solution is valid.

So, we have that |x+3|=4x-3 has only one root: x=2.

Hope it's clear.

Why do we discard that solution?
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Re: [tricky one] Is x > 0? [#permalink]  04 Jun 2013, 08:13
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Expert's post
WholeLottaLove wrote:
Bunuel wrote:
|x+3|=4x-3. Check point is at x=-3 (check point, is the value of x for which the value of an expression in modulus equals to zero).

When x\leq{-3}, then x+3<0, thus |x+3|=-(x+3). So, in this case we have -(x+3)=4x-3 --> x=0 --> discard this solution since we consider the range when x\leq{-3}.

When x>{-3}, then x+3>0, thus |x+3|=x+3. So, in this case we have x+3=4x-3 --> x=2 --> since x=2>-3, then this solution is valid.

So, we have that |x+3|=4x-3 has only one root: x=2.

Hope it's clear.

Why do we discard that solution?

We consider the range x\leq{-3}. x=0 is out of this range.

Check absolute values chapter of math book: math-absolute-value-modulus-86462.html

Hope it helps.
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Re: [tricky one] Is x > 0? [#permalink]  04 Jun 2013, 10:25
I think I got it. Thanks!

Bunuel wrote:
WholeLottaLove wrote:
Bunuel wrote:
|x+3|=4x-3. Check point is at x=-3 (check point, is the value of x for which the value of an expression in modulus equals to zero).

When x\leq{-3}, then x+3<0, thus |x+3|=-(x+3). So, in this case we have -(x+3)=4x-3 --> x=0 --> discard this solution since we consider the range when x\leq{-3}.

When x>{-3}, then x+3>0, thus |x+3|=x+3. So, in this case we have x+3=4x-3 --> x=2 --> since x=2>-3, then this solution is valid.

So, we have that |x+3|=4x-3 has only one root: x=2.

Hope it's clear.

Why do we discard that solution?

We consider the range x\leq{-3}. x=0 is out of this range.

Check absolute values chapter of math book: math-absolute-value-modulus-86462.html

Hope it helps.
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]  17 Jun 2013, 22:45
Hi Bunnel,
I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ...

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard[Understood this part]2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution )[Did not understand, Why -ve sign is here as the x range is positive]3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)[Understood]
Another thing is 3/2(inclusive) is not at all considered in the range in any of the three checkpoints.

What am i missing here and how to solve these questions effectively
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]  17 Jun 2013, 23:50
Expert's post
kawan84 wrote:
Hi Bunnel,
I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ...

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard[Understood this part]2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution )[Did not understand, Why -ve sign is here as the x range is positive]3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)[Understood]
Another thing is 3/2(inclusive) is not at all considered in the range in any of the three checkpoints.

What am i missing here and how to solve these questions effectively

Absolute value properties:

When x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|={-(some \ expression)}. For example: |-5|=5=-(-5);

When x\geq{0} then |x|=x, or more generally when some \ expression\geq{0} then |some \ expression|={some \ expression}. For example: |5|=5.

SOLUTION:

We have two transition points for |x-3|=|2x-3|: x=\frac{3}{2} and x=3. Thus three ranges to check:
1. x<\frac{3}{2};
2. \frac{3}{2}\leq{x}\leq{3};
3. 3<x

Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.

1. When x<\frac{3}{2}, then x-3 is negative and 2x-3 is negative too, thus |x-3|=-(x-3) and |2x-3|=-(2x-3).

Therefore for this range |x-3|=|2x-3| transforms to -(x-3)=-(2x-3) --> x=0. This solution is OK, since x=0 is in the range we consider (x<\frac{3}{2}).

2. When \frac{3}{2}\leq{x}\leq{3}, then x-3 is negative and 2x-3 is positive, thus |x-3|=-(x-3) and |2x-3|=2x-3.

Therefore for this range |x-3|=|2x-3| transforms to -(x-3)=2x-3 --> x=2. This solution is OK, since x=2 is in the range we consider (\frac{3}{2}\leq{x}\leq{3}).

3. When 3<x, then x-3 is positive and 2x-3 is positive too, thus |x-3|=x-3 and |2x-3|=2x-3.

Therefore for this range |x-3|=|2x-3| transforms to x-3=2x-3 --> x=0. This solution is NOT OK, since x=0 is NOT in the range we consider (3<x).

Thus |x-3|=|2x-3| has two solutions x=0 and x=2.

Hope it's clear.

P.S. Though for this particular question I still suggest another approach shown in my post here: is-x-0-1-x-3-4x-3-2-x-3-2x-127978.html#p1048512
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Re: [tricky one] Is x > 0? [#permalink]  21 Jun 2013, 23:40
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. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) |x-3|=|2x-3|. Square both sides: (x-3)^2=(2x-3)^2 --> (2x-3)^2-(x-3)^2=0. Apply a^2-b^2=(a-b)(a+b), rather than squaring: --> x(3x-6)=0 --> x=0 or x=2. Not sufficient.

Hope it helps.

For statement 1,
let us square both sides of the equation and remove the modulus sign.

|x+3|=4x-3
squaring both sides we get,
(|x+3|)^2 = (4x-3)^2
= x^2 + 6x + 9 = 16x^2 - 24x + 9
= 15x^2 - 30x = 0
x^2 - 2x = 0

x can be 0 or 2.
Hence Statement 1 is not sufficient.
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Re: [tricky one] Is x > 0? [#permalink]  22 Jun 2013, 02:43
Expert's post
navigator123 wrote:
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. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) |x-3|=|2x-3|. Square both sides: (x-3)^2=(2x-3)^2 --> (2x-3)^2-(x-3)^2=0. Apply a^2-b^2=(a-b)(a+b), rather than squaring: --> x(3x-6)=0 --> x=0 or x=2. Not sufficient.

Hope it helps.

For statement 1,
let us square both sides of the equation and remove the modulus sign.

|x+3|=4x-3
squaring both sides we get,
(|x+3|)^2 = (4x-3)^2
= x^2 + 6x + 9 = 16x^2 - 24x + 9
= 15x^2 - 30x = 0
x^2 - 2x = 0

x can be 0 or 2.
Hence Statement 1 is not sufficient.

Substitute 0 in the equation, does it hold true?
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]  27 Jun 2013, 08:02
Is x > 0?

(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|

(1) |x + 3| = 4x – 3

Well, we know that (4x-3) must be greater than or equal to zero: 4x-3 >= 0 ===> 4x>=3 ===> x>=3/4

So, for #1 I would say that it is sufficient. For 4x-3 to be positive x must be greater than or equal to 3/4.
SUFFICIENT

(2) |x – 3| = |2x – 3|

We don't know if both sides are positive so we can't take the square root. So: |x – 3| = |2x – 3|

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

Here, we have two valid solutions.
INSUFFICIENT

EDIT: Apparently you CAN square both sides in #2. I thought you had to know that, for example (x-3) and (2x-3) are positive in order for you to be able to square both sides?
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Re: [tricky one] Is x > 0? [#permalink]  27 Jul 2013, 09:54
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. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) |x-3|=|2x-3|. Square both sides: (x-3)^2=(2x-3)^2 --> (2x-3)^2-(x-3)^2=0. Apply a^2-b^2=(a-b)(a+b), rather than squaring: --> x(3x-6)=0 --> x=0 or x=2. Not sufficient.

Hope it helps.

for (2) why do you start by squaring both sides? What is the rule that you are using to determine that this is the first step?

Thank you!
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Re: [tricky one] Is x > 0? [#permalink]  27 Jul 2013, 13:06
You can square both sides of an inequality or equation when there are absolute values involved. Think of it like this:

x^2 = y^2
|x| = |y|

Lets say x^2 = 4. Now, x can be 2 OR -2 because squaring any number positive or negative will result in a number ≥ 0.

kanderoo wrote:
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. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) |x-3|=|2x-3|. Square both sides: (x-3)^2=(2x-3)^2 --> (2x-3)^2-(x-3)^2=0. Apply a^2-b^2=(a-b)(a+b), rather than squaring: --> x(3x-6)=0 --> x=0 or x=2. Not sufficient.

Hope it helps.

for (2) why do you start by squaring both sides? What is the rule that you are using to determine that this is the first step?

Thank you!
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]  05 Nov 2013, 12:00
The original post is incorrect 2) should read (2) |x + 1| = 2x – 1

edit: If the OP is supposed to be from manhattan
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Re: [tricky one] Is x > 0? [#permalink]  06 Nov 2013, 08:43
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. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) |x-3|=|2x-3|. Square both sides: (x-3)^2=(2x-3)^2 --> (2x-3)^2-(x-3)^2=0. Apply a^2-b^2=(a-b)(a+b), rather than squaring: --> x(3x-6)=0 --> x=0 or x=2. Not sufficient.

Hope it helps.

Sneaky problem. That's interesting that you're expected to catch this before solving the equation.

I solved for x=2 and x=0 in A and then tested to find that 0 wasn't actually valid (repeating the same process in B to find that A was correct answer).

Also, could you explain again why you will sometimes solve for answers in these absolute value equations that aren't actually valid solutions? That doesn't make sense to me.
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Re: [tricky one] Is x > 0? [#permalink]  07 Nov 2013, 04:56
Expert's post
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. You can see here that you don't even need to find exact value(s) of x to answer the question.

(2) |x-3|=|2x-3|. Square both sides: (x-3)^2=(2x-3)^2 --> (2x-3)^2-(x-3)^2=0. Apply a^2-b^2=(a-b)(a+b), rather than squaring: --> x(3x-6)=0 --> x=0 or x=2. Not sufficient.

Hope it helps.

Sneaky problem. That's interesting that you're expected to catch this before solving the equation.

I solved for x=2 and x=0 in A and then tested to find that 0 wasn't actually valid (repeating the same process in B to find that A was correct answer).

Also, could you explain again why you will sometimes solve for answers in these absolute value equations that aren't actually valid solutions? That doesn't make sense to me.

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Re: [tricky one] Is x > 0?   [#permalink] 07 Nov 2013, 04:56
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