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Is x > 0? (1) |10 + 3| = 4x – 3 (2) |x – three| = |2x – 3| [#permalink]   22 February 2012, xviii:59

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Is x > 0?

(1) |x + iii| = 4x – three
(ii) |ten – three| = |2x – 3|

Math Expert

Joined: 02 Sep 2009

Posts: 86968

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]   22 Feb 2012, 21:02

Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an accented value, which is always non-negative (\(|some \ expression|\geq{0}\)), then RHS must also be not-negative --> \(4x-iii\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can see here that y'all don't fifty-fifty need to observe exact value(s) of x to answer the question.

(2) \(|10-three|=|2x-3|\). Square both sides: \((x-three)^2=(2x-three)^2\) --> \((2x-3)^2-(x-three)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(ten=two\). Not sufficient.

Answer: A.

Hope it helps.
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Re: Is ten > 0? (1) |x + iii| = 4x – 3 (two) |x – 3| = |2x – 3| [#permalink]   25 Feb 2012, 09:45

Bunnel,
For statement ane, I understand that you accept not solved for x in your solution. I am withal learning the abs value and in equalities. Tin y'all share your thoughts on the following solution (solving for x in both statements)

a. |ten+3| = 4x -three
|ten+3| = x + 3 for x + 3 >= 0 ---> 10 > -3.
Solving for x,
ten + 3 = 4x -3 ---> half dozen = 3x ---> ten = 2 ( solution accepted since 10 > -3)

| x + 3| = -(x +3 ) for 10 + 3 < 0 --> x < -3.
Solving for x,
-x - 3 = 4x -3
5x = 0 ---> 10 = 0 ( solution discarded as 10 is not < -3)

A is sufficient.

b. |x -3| = |2x -three|
|x -3| = ten -3 for x-3 >= 0 ---> x >= 3
|ten -3| = -(10 -three) for x-3 < 0 ---> x < iii

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

So nosotros accept 3 ranges

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard
2) 3/2 < x < three ===> -(10-three) = 2x -three ====> -10 + iii = 2x -3 ===> 10 =two ( accepted solution )
3) x < iii/ii ===> -(x-iii) = -(2x -iii) ====> x = three (discarded the solution)

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

One more than question, if we consider the only acceptable solution in argument b i.e., x =2 (case iii/ii < ten < 3), then both statements are sufficient independently and we get the answer as option (D).

Please discuss.

Regards

Math Expert

Joined: 02 Sep 2009

Posts: 86968

Re: Is x > 0? (1) |ten + iii| = 4x – iii (2) |10 – 3| = |2x – 3| [#permalink]   25 Feb 2012, 09:52

saxenaashi wrote:

Bunnel,
For statement i, I understand that you take not solved for 10 in your solution. I am withal learning the abs value and in equalities. Tin you share your thoughts on the following solution (solving for x in both statements)

a. |10+iii| = 4x -3
|10+3| = ten + 3 for x + 3 >= 0 ---> x > -3.
Solving for x,
10 + 3 = 4x -3 ---> 6 = 3x ---> x = 2 ( solution accepted since 10 > -3)

| 10 + 3| = -(x +three ) for 10 + 3 < 0 --> ten < -3.
Solving for ten,
-x - 3 = 4x -3
5x = 0 ---> 10 = 0 ( solution discarded equally x is not < -iii)

A is sufficient.

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

|2x -iii | = 2x -iii for 2x -iii >= 0 ----> ten >=iii/ii
|2x -3 | = -(2x -3) for 2x -three < 0 ----> ten < 3/ii

And so nosotros 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 ===> 10 =2 ( accepted solution )
iii) 10 < 3/two ===> -(10-3) = -(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 nosotros consider the merely acceptable solution in statement b i.east., x =ii (case 3/ii < x < three), and so both statements are sufficient independently and we become the answer as option (D).

Please talk over.

Regards

Information technology seems that you lot empathize this method very well.

Everything is correct except the red parts: -(x-iii) = -(2x -3) --> 10=0 (not x=3), so it's besides a valid solution. Hence for (ii) you accept two valid solutions x=2 and ten=0 (just similar in my post in a higher place), which makes this argument not sufficient.

Promise it'south articulate.
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Re: Is x > 0? (1) |x + three| = 4x – iii (two) |x – 3| = |2x – iii| [#permalink]   25 Feb 2012, x:07

Bunuel wrote:

saxenaashi wrote:

Bunnel,
For argument one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Tin you share your thoughts on the following solution (solving for x in both statements)

a. |x+3| = 4x -3
|x+3| = x + 3 for 10 + iii >= 0 ---> x > -iii.
Solving for ten,
ten + three = 4x -3 ---> 6 = 3x ---> x = ii ( solution accepted since x > -iii)

| ten + three| = -(ten +3 ) for x + 3 < 0 --> x < -three.
Solving for x,
-x - iii = 4x -iii
5x = 0 ---> x = 0 ( solution discarded equally 10 is not < -3)

A is sufficient.

b. |x -3| = |2x -3|
|x -3| = x -three for x-3 >= 0 ---> ten >= 3
|ten -iii| = -(x -3) for 10-3 < 0 ---> x < iii

|2x -3 | = 2x -3 for 2x -3 >= 0 ----> x >=3/2
|2x -3 | = -(2x -3) for 2x -3 < 0 ----> 10 < 3/2

So nosotros take 3 ranges

ane) x >=3 ==> x - three = 2x -three ====> x = 0 ; discard
2) 3/ii < 10 < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> 10 =2 ( accepted solution )
3) ten < 3/2 ===> -(ten-3) = -(2x -3) ====> x = three (discarded the solution)

Allow me know if I take solved the question correctly. I know the process is lengthy.

One more question, if we consider the only acceptable solution in statement b i.due east., x =ii (case 3/2 < x < 3), and then both statements are sufficient independently and we get the reply as option (D).

Please discuss.

Regards

Information technology seems that you understand this method very well.

Everything is correct except the red parts: -(x-3) = -(2x -3) --> 10=0 (not x=3), so it's as well a valid solution. Hence for (ii) you accept two valid solutions 10=2 and ten=0 (just like in my postal service higher up), which makes this statement not sufficient.

Hope it'southward clear.

Works. Thanks for pointing out the error and quick response likewise. I am just learning this and so working ground upward, hence I am not hesitant in taking long route initially. After adjacent few questions, I would be tuned to extract the ranges I guess. Is there a condition on squaring the sides of the equation or inequation.

Regards

Math Adept

Joined: 02 Sep 2009

Posts: 86968

Re: Is 10 > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – iii| [#permalink]   25 Feb 2012, 10:xv

saxenaashi wrote:

Works. Thanks for pointing out the fault and quick response too. I am merely learning this so working ground upward, hence I am not hesitant in taking long road initially. After next few questions, I would be tuned to excerpt the ranges I guess. Is there a condition on squaring the sides of the equation or inequation.

Regards

A. We tin can raise both parts of an inequality to an fifty-fifty power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(two<4\) --> we can foursquare both sides and write: \(2^2<iv^2\);
\(0\leq{10}<{y}\) --> we can square both sides and write: \(x^two<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. And so if given that \(x>y\) then nosotros can non square both sides and write \(10^2>y^ii\) if we are non certain that both \(10\) and \(y\) are non-negative.

B. We can e'er raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> nosotros can raise both sides to tertiary power and write: \(-2^3=-viii<-one=-one^3\) or \(-v<1\) --> \(-5^ii=-125<ane=1^iii\);
\(10<y\) --> we can enhance both sides to third power and write: \(x^three<y^three\).

So in statement (2) since both parts of expression are non-negative we tin safely apply squaring.

Hope information technology helps.
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Re: Is x > 0? (one) |x + 3| = 4x – 3 (two) |10 – 3| = |2x – three| [#permalink]   27 Feb 2012, 22:09

Well, I have a little bit another solution for this trouble

Consider (i): if ten>-3 then x+3=4x-3, so x=2, sufficient
if x<-3, then -x-iii=4x-3, which Leeds to respond ten=0, which is wrong because we consider only 10<-3
Sum upward (1) is sufficient

The saim logic can be practical to the (ii), and information technology is possible to derive that (2) is sufficient

And then the answer is D either (1) is sufficient or (2) is sufficient

In your logic, yous gave a mistake, because you've missed the root.

Posted from GMAT ToolKit

Math Expert

Joined: 02 Sep 2009

Posts: 86968

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |ten – three| = |2x – iii| [#permalink]   27 Feb 2012, 22:thirty

terance wrote:

Well, I take a little bit another solution for this trouble

Consider (1): if ten>-iii then x+iii=4x-iii, then x=2, sufficient
if x<-iii, so -x-three=4x-three, which Leeds to respond 10=0, which is wrong because we consider just ten<-3
Sum up (ane) is sufficient

The saim logic can exist applied to the (2), and it is possible to derive that (2) is sufficient

So the answer is D either (ane) is sufficient or (2) is sufficient

In your logic, y'all gave a fault, because you lot've missed the root.

Posted from GMAT ToolKit

Welcome to GMAT Club.

Unfortunately your answer is non correct: OA for this question is A, not D (you tin see it under the spoiler in the initial mail).

(2) |ten – 3| = |2x – 3| has two roots 10=0 and ten=two (just substitute these values to run across that they both satisfy the given equation), so you can not get the single numerical value of 10, which makes this statement insufficient.

Y'all can refer to above solutions for two unlike approaches of how to get these roots for (ii). Please ask if anything remains unclear.
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Re: Is 10 > 0? (ane) |x + 3| = 4x – 3 (2) |10 – 3| = |2x – 3| [#permalink]   17 April 2012, 05:xix

dvinoth86 wrote:

Is 10 > 0?

(1) |ten + three| = 4x – iii
(ii) |x – three| = |2x – three|

Liked the question? encourage by giving kudos

Recollect: When you have || on both sides of eqn y'all do not need to VERIFY the answer past putting them back in eqn Simply when y'all have || on only one side y'all MUST VERIFY the answer past putting them back in eqn

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

Hence Sufficient

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

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Posts: 52

Re: Is x > 0? (i) |x + 3| = 4x – three (two) |10 – 3| = |2x – three| [#permalink]   09 May 2013, 12:38

Hi Bunuel
This is how I solved the ii statements,

http://campl.u.s./fWdgqiT5EsK

and I go 0 & 2 for both statements, tin can u tell me what wrong am I doing?

Posted from my mobile device

Math Good

Joined: 02 Sep 2009

Posts: 86968

Re: Is x > 0? (1) |ten + iii| = 4x – iii (2) |ten – 3| = |2x – 3| [#permalink]   10 May 2013, 01:01

nikhil007 wrote:

Hi Bunuel
This is how I solved the two statements,

http://campl.us/fWdgqiT5EsK

and I get 0 & 2 for both statements, tin u tell me what wrong am I doing?

Posted from my mobile device

If y'all substitute x=0 in |x + 3| = 4x – 3 you'll get: LHS=|x + 3|=3 and RHS=4x – iii=-3, thus \(LHS\neq{RHS}\), which means that x=0 is non the root of the given equation.

When expanding |x+3|:

When x<-iii, then |x+3|=-(10+3), and then in this case we'll have -(ten+iii)=4x-3 --> x=0 --> discard this value since 0 is not less than -3 (we consider the range when x<-3).

When x>=-3, then |x+3|=x+3, so in this case we'll take x+3=4x-3 --> x=2 --> this value of x is OK since two>-3.

So, |x + 3| = 4x - 3 has only ane root, x=ii.

Hope it'due south articulate.
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Posts: 324

Re: Is x > 0? (1) |x + iii| = 4x – 3 (2) |ten – 3| = |2x – 3| [#permalink]   04 Jun 2013, 06:56

Hi!

Why do nosotros non find the positive and negative values for |x+three|=4x-iii? Is it because this isn't a <, >, >= problem?

Thanks!

Bunuel wrote:

Is x > 0?

(one) \(|10+3|=4x-3\) --> LHS is an accented value, which is ever non-negative (\(|some \ expression|\geq{0}\)), and so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{iv}}\), hence \(ten>0\). Sufficient. You can see here that you don't even need to observe verbal value(s) of ten to answer the question.

(two) \(|10-iii|=|2x-iii|\). Square both sides: \((x-3)^ii=(2x-iii)^2\) --> \((2x-iii)^2-(10-3)^2=0\). Apply \(a^2-b^ii=(a-b)(a+b)\), rather than squaring: --> \(x(3x-six)=0\) --> \(x=0\) or \(10=two\). Not sufficient.

Answer: A.

Promise it helps.

Math Expert

Joined: 02 Sep 2009

Posts: 86968

Re: Is x > 0? (1) |x + iii| = 4x – three (ii) |x – 3| = |2x – 3| [#permalink]   04 Jun 2013, 07:00

WholeLottaLove wrote:

Hi!

Why exercise nosotros not find the positive and negative values for |x+three|=4x-3? Is it because this isn't a <, >, >= problem?

Cheers!

Bunuel wrote:

Is x > 0?

(i) \(|10+iii|=4x-iii\) --> LHS is an absolute value, which is e'er non-negative (\(|some \ expression|\geq{0}\)), so RHS must likewise be non-negative --> \(4x-iii\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. You can run into hither that you don't even demand to find verbal value(s) of x to reply the question.

(2) \(|x-3|=|2x-iii|\). Foursquare both sides: \((10-3)^two=(2x-iii)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(10=0\) or \(x=2\). Non sufficient.

Respond: A.

Hope information technology helps.

I tin do this way besides, simply the manner shown in my post is faster.
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Posts: 324

Re: Is ten > 0? (i) |ten + three| = 4x – 3 (2) |ten – three| = |2x – 3| [#permalink]   04 Jun 2013, 07:26

Hmmm...

When I solve for the pos. and neg. values of |ten+3|=4x-3 I get:

I. 10+3=4x-3 ==> -3x=-half-dozen ==> x=2
2. x+3=-(4x-3) ==> x+3=-4x+3 ==> 5x=0 ==> ten=0

And then hither in my presumably incorrect simplification, I have x=ii and x=0 in which example we can't exist certain if x>0

For many abs. value questions it seems that yous accept to find the positive and negative cases each equation. I get that in this instance, |x+3|=4x-3 means that 4x-3 is positive merely why for other, similar questions, is solving for the positive and negative cases necessary?

Bunuel wrote:

WholeLottaLove wrote:

Hi!

Why do we not find the positive and negative values for |ten+3|=4x-3? Is it because this isn't a <, >, >= problem?

Thank you!

Bunuel wrote:

Is x > 0?

(1) \(|x+three|=4x-iii\) --> LHS is an absolute value, which is ever non-negative (\(|some \ expression|\geq{0}\)), so RHS must besides be not-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient. Yous tin run into here that you don't even need to observe exact value(s) of x to answer the question.

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^ii=(2x-3)^2\) --> \((2x-3)^ii-(x-iii)^ii=0\). 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.

One can do this way too, but the style shown in my mail service is faster.

Math Practiced

Joined: 02 Sep 2009

Posts: 86968

Re: Is 10 > 0? (i) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]   04 Jun 2013, 07:42

\(|x+iii|=4x-3\). Check point is at \(ten=-3\) (check bespeak, is the value of x for which the value of an expression in modulus equals to zero).

When \(10\leq{-3}\), and then \(x+3<0\), thus \(|x+3|=-(x+3)\). And so, in this example we have \(-(ten+3)=4x-3\) --> \(ten=0\) --> discard this solution since nosotros consider the range when \(x\leq{-three}\).

When \(x>{-three}\), then \(x+3>0\), thus \(|x+3|=ten+3\). And then, in this case we have \(ten+three=4x-3\) --> \(10=two\) --> since \(x=2>-3\), so this solution is valid.

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

Hope information technology'southward articulate.
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Re: Is x > 0? (1) |10 + iii| = 4x – iii (2) |ten – 3| = |2x – 3| [#permalink]   04 Jun 2013, 08:10

Bunuel wrote:

\(|10+3|=4x-3\). Check betoken is at \(x=-three\) (check point, is the value of x for which the value of an expression in modulus equals to zero).

When \(x\leq{-3}\), then \(ten+3<0\), thus \(|x+3|=-(10+3)\). Then, in this case nosotros have \(-(ten+3)=4x-iii\) --> \(ten=0\) --> discard this solution since we consider the range when \(x\leq{-3}\).

When \(x>{-3}\), then \(ten+3>0\), thus \(|x+3|=10+3\). And then, in this case nosotros have \(x+3=4x-3\) --> \(x=2\) --> since \(x=2>-3\), and then this solution is valid.

So, we take that \(|x+three|=4x-three\) has merely i root: \(ten=2\).

Hope it'southward clear.

Why do we discard that solution?

Math Expert

Joined: 02 Sep 2009

Posts: 86968

Re: Is x > 0? (1) |10 + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]   04 Jun 2013, 08:13

WholeLottaLove wrote:

Bunuel wrote:

\(|x+3|=4x-three\). Bank check bespeak is at \(x=-3\) (cheque point, is the value of ten for which the value of an expression in modulus equals to zero).

When \(x\leq{-3}\), then \(ten+3<0\), thus \(|10+3|=-(x+3)\). And then, in this case we accept \(-(x+3)=4x-3\) --> \(x=0\) --> discard this solution since we consider the range when \(x\leq{-3}\).

When \(x>{-three}\), then \(x+3>0\), thus \(|x+3|=x+3\). Then, in this case we accept \(ten+three=4x-three\) --> \(x=2\) --> since \(x=two>-3\), then this solution is valid.

So, we accept that \(|x+3|=4x-three\) has only one root: \(10=two\).

Promise information technology'south clear.

Why do we discard that solution?

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

Bank check absolute values affiliate of math book:

math-absolute-value-modulus-86462.html

Hope it helps.
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Re: Is x > 0? (1) |x + iii| = 4x – iii (two) |x – 3| = |2x – iii| [#permalink]   17 Jun 2013, 22:45

Hullo Bunnel,
I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ...

1) x >=3 ==> ten - iii = 2x -iii ====> x = 0 ; discard[Understood this part]2) iii/ii < ten < three ===> -(x-3) = 2x -three ====> -x + three = 2x -3 ===> x =two ( accepted solution )[Did not empathise, Why -ve sign is here as the ten range is positive]3) x < 3/2 ===> -(ten-3) = -(2x -3) ====> 10 = iii (discarded the solution)[Understood]
Some other affair 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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Joined: 02 Sep 2009

Posts: 86968

Re: Is x > 0? (1) |x + iii| = 4x – iii (ii) |x – 3| = |2x – 3| [#permalink]   17 Jun 2013, 23:50

kawan84 wrote:

Hi Bunnel,
I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ...

1) x >=three ==> ten - 3 = 2x -3 ====> x = 0 ; discard[Understood this part]2) iii/2 < ten < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -three ===> 10 =2 ( accepted solution )[Did not understand, Why -ve sign is hither as the 10 range is positive]3) x < iii/2 ===> -(x-iii) = -(2x -3) ====> x = 3 (discarded the solution)[Understood]
Another matter is iii/two(inclusive) is not at all considered in the range in whatever of the three checkpoints.

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

Absolute value properties:

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

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

SOLUTION:

We have ii transition points for \(|x-iii|=|2x-3|\): \(x=\frac{iii}{two}\) and \(x=3\). Thus three ranges to cheque:
1. \(x<\frac{3}{two}\);
2. \(\frac{3}{2}\leq{ten}\leq{three}\);
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.

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

Therefore for this range \(|10-iii|=|2x-3|\) transforms to \(-(ten-3)=-(2x-3)\) --> \(10=0\). This solution is OK, since \(x=0\) is in the range we consider (\(x<\frac{iii}{2}\)).

ii. When \(\frac{3}{2}\leq{10}\leq{3}\), then \(x-3\) is negative and \(2x-iii\) is positive, thus \(|10-3|=-(x-3)\) and \(|2x-3|=2x-3\).

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

3. When \(three<x\), then \(10-3\) is positive and \(2x-three\) is positive also, thus \(|ten-3|=x-3\) and \(|2x-3|=2x-3\).

Therefore for this range \(|10-3|=|2x-3|\) transforms to \(ten-3=2x-three\) --> \(x=0\). This solution is NOT OK, since \(10=0\) is Non in the range we consider (\(3<ten\)).

Thus \(|x-three|=|2x-3|\) has ii solutions \(10=0\) and \(x=2\).

Hope it'south clear.

P.S. Though for this particular question I yet advise another arroyo 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: Is ten > 0? (1) |x + 3| = 4x – iii (two) |x – iii| = |2x – 3| [#permalink]   27 Jun 2013, 08:02

Is 10 > 0?

(one) |10 + iii| = 4x – 3
(2) |ten – iii| = |2x – 3|

(1) |x + 3| = 4x – iii

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

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

(2) |x – 3| = |2x – 3|

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

(x-3)=(2x-3) ===> x-3=2x-three ===> 3x=0 ===> x=0
OR
(x-three)=-2x+3 ===> x-three=-2x+three ===> 3x=6 ===> 10=ii

Hither, we have two valid solutions.
Bereft

EDIT: Obviously y'all Can square both sides in #2. I idea you had to know that, for instance (x-3) and (2x-three) are positive in order for y'all to exist able to square both sides?

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

27 Jun 2013, 08:02

4x 3 X 2x 3,

Source: https://gmatclub.com/forum/is-x-0-1-x-3-4x-3-2-x-3-2x-127978.html

Posted by: gonzalesmiled1958.blogspot.com

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