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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]

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04 Jun 2013, 08:42
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$$|x+3|=4x-3$$. 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]

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04 Jun 2013, 09:10
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]

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04 Jun 2013, 09:13
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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]

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04 Jun 2013, 11:25
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]

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17 Jun 2013, 23:45
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]

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18 Jun 2013, 00:50
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]

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22 Jun 2013, 00:40
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.

Answer: A.

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.
Please correct me if i had made a mistake.
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Re: [tricky one] Is x > 0? [#permalink]

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22 Jun 2013, 03:43
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.

Answer: A.

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.
Please correct me if i had made a mistake.

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]

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27 Jun 2013, 09:02
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]

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27 Jul 2013, 10:54
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.

Answer: A.

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]

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27 Jul 2013, 14:06
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.

Answer: A.

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]

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06 Nov 2013, 09:43
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.

Answer: A.

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]

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07 Nov 2013, 05:56
vandygrad11 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.

Answer: A.

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.

It depends on how you solve. So, please show your work.
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]

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16 Apr 2014, 07:23
a. |x+3| = 4x -3
|x+3| = x + 3 for x + 3 >= 0 ---> x > -3.
Solving for x,
x + 3 = 4x -3 ---> 6 = 3x ---> x = 2 ( solution accepted since x > -3)

| x + 3| = -(x +3 ) for x + 3 < 0 --> x < -3.
Solving for x,
-x - 3 = 4x -3
5x = 0 ---> x = 0 ( solution discarded as x is not < -3)

A is sufficient.

b. |x -3| = |2x -3|
|x -3| = x -3 for x-3 >= 0 ---> x >= 3
|x -3| = -(x -3) for x-3 < 0 ---> x < 3

|2x -3 | = 2x -3 for 2x -3 >= 0 ----> x >=3/2
|2x -3 | = -(2x -3) for 2x -3 < 0 ----> x < 3/2

So we have 3 ranges

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard
2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution )
3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)

Let me know if I have solved the question correctly. I know the process is lengthy.

One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D).
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]

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16 Apr 2014, 07:23
a. |x+3| = 4x -3
|x+3| = x + 3 for x + 3 >= 0 ---> x > -3.
Solving for x,
x + 3 = 4x -3 ---> 6 = 3x ---> x = 2 ( solution accepted since x > -3)

| x + 3| = -(x +3 ) for x + 3 < 0 --> x < -3.
Solving for x,
-x - 3 = 4x -3
5x = 0 ---> x = 0 ( solution discarded as x is not < -3)

A is sufficient.

b. |x -3| = |2x -3|
|x -3| = x -3 for x-3 >= 0 ---> x >= 3
|x -3| = -(x -3) for x-3 < 0 ---> x < 3

|2x -3 | = 2x -3 for 2x -3 >= 0 ----> x >=3/2
|2x -3 | = -(2x -3) for 2x -3 < 0 ----> x < 3/2

So we have 3 ranges

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard
2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution )
3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)

Let me know if I have solved the question correctly. I know the process is lengthy.

One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D).
tnx for this
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Is x > 0 ? (Inequalities and absolute values) [#permalink]

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19 May 2014, 09:02
Is x > 0?

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

Please provide your valuable explanations.
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Re: Is x > 0 ? (Inequalities and absolute values) [#permalink]

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19 May 2014, 09:06
DesiGmat wrote:
Is x > 0?

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

Please provide your valuable explanations.

Merging similar topics. Please refer to the discussion on pages 1 and 2.
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Re: Is x > 0 ? (Inequalities and absolute values) [#permalink]

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19 May 2014, 20:37
DesiGmat wrote:
Is x > 0?

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

Please provide your valuable explanations.

It has already been answered above. but for your convenience:

Statement 1 says:

|x + 3| = 4x – 3

Now since |x + 3|>=0 always, 4x – 3 >=0 as well

x>=3/4 >0 . Hence, it is sufficient.

Statement 2:

|x – 3| = |2x – 3|

Since both are positive numbers, we can square them and find x which is coming out to be 0 and 2. Thus, it is not sufficient.

Answer therefore is A

Hope you understand

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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]

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12 May 2015, 11:56
vikram4689 wrote:
dvinoth86 wrote:
Is x > 0?

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

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Remember: When you have || on both sides of eqn you do not need to VERIFY the answer by putting them back in eqn BUT when you have || on only one side you MUST VERIFY the answer by putting them back in eqn

(1) |x + 3| = 4x – 3
a) x + 3 = 4x – 3 => x=2 .. VALID
b) -(x + 3) = 4x – 3 => x=0 .. INVALID

Hence Sufficient

(2) |x – 3| = |2x – 3| => x = 0 or 6 .. INVALID
Hence In-Sufficient

Your statement 1 has two sceneries. Then how does it become sufficient?
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]

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12 May 2015, 23:57
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ranaazad wrote:
vikram4689 wrote:
dvinoth86 wrote:
Is x > 0?

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

Liked the question? encourage by giving kudos

Remember: When you have || on both sides of eqn you do not need to VERIFY the answer by putting them back in eqn BUT when you have || on only one side you MUST VERIFY the answer by putting them back in eqn

(1) |x + 3| = 4x – 3
a) x + 3 = 4x – 3 => x=2 .. VALID
b) -(x + 3) = 4x – 3 => x=0 .. INVALID

Hence Sufficient

(2) |x – 3| = |2x – 3| => x = 0 or 6 .. INVALID
Hence In-Sufficient

Your statement 1 has two sceneries. Then how does it become sufficient?

Dear ranaazad

The solution quoted by you above has some tacit steps. I'll list them here:

St. 1 says: |x + 3| = 4x – 3 . . . (1)

Case 1: x + 3 > = 0

That is, x > = -3

In this case, |x+3| = (x+3)

So, Equation 1 becomes:

x + 3 = 4x - 3
=> 6 = 3x
=> x = 2

Does this value of x satisfy the condition of Case 1, that x > = -3?

Yes, it does. So, x = 2 is a valid value of x.

Let's now consider

Case 2: x + 3 < 0

That is, x < -3

In this case, |x+3| = -(x+3)

So, Equation 1 becomes:

-(x + 3) = 4x - 3
=> -x - 3 = 4x - 3
=> 0 = 5x
=> x = 0

Does this value of x satisfy the condition of Case 2, that x < -3?

No, it doesn't. So, x = 0 is an INVALID value of x.

Thus, we got only 1 valid value of X from Statement 1 (x = 2) and so, Statement 1 is sufficient to say that x > 0.

Hope this answered your doubt!

Japinder
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|   [#permalink] 12 May 2015, 23:57

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