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Re: Inequality - absolute on both sides [#permalink]
28 Jul 2010, 03:22
4
This post received KUDOS
Expert's post
5
This post was BOOKMARKED
mn2010 wrote:
The question is
Is |x| < 1 ?
(1) |x + 1| = 2|x – 1|
(2) |x – 3| > 0
I always get confused how to approach the equality or inequality with absolute signs on both sides (statement 1). Any1 knows an efficient way to approach this ?
Is \(|x| < 1\)?
Is \(|x| < 1\), means is \(x\) in the range (-1,1) or is \(-1<x<1\) true?
(1) \(|x + 1| = 2|x - 1|\) Two key points: \(x=-1\) and \(x=1\) (key points are the values of \(x\) when absolute values equal to zero), thus three ranges to check: ---------{-1}--------{1}---------
A. \(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\) --> \(x=3\), not OK, as this value is not in the range we are checking (\(x<-1\)); B. \(-1\leq{x}\leq{1}\) (green range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\). OK, as this value is in the range we are checking (\(-1\leq{x}\leq{1}\)); C. \(x>1\) (red range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(x-1)\) --> \(x=3\). OK, as this value is in the range we are checking (\(x>1\)).
So we got TWO values of \(x\) (two solutions): \(\frac{1}{3}\) and \(3\), first is in the range (-1,1) but second is out of the range. Not sufficient.
Alternately we could just expand absolute values of RHS and LHS with same sign (for example both LHS and RHS positive: \(x + 1 = 2(x - 1)\)) and then with different sign (for example LHS positive and RHS negative \(x + 1 = 2(-x + 1)\)), solve for \(x\) both equations, and finally check whether the solutions satisfy \(|x + 1| = 2|x - 1|\).
OR: we can square given equation to get rid of the modulus: \((x + 1)^2 = 4(x - 1)^2\) --> \(3x^2-10x+3=0\) --> \(x=3\) or \(x=\frac{1}{3}\).
(2) \(|x - 3|>{0}\). Absolute value is always non-negative, more than or equal to zero: \(|some \ expression|\geq{0}\). We are told that absolute value of \(x-3\) is MORE than zero, so just it says that \(|x-3|\neq{0}\), which simply means that \(x\neq{3}\). But we don't know whether \(x\) is in the range (-1,1) or not.
(1)+(2) \(x=\frac{1}{3}\) or \(x=3\) AND \(x\neq{3}\) --> means \(x\) can have only value \(\frac{1}{3}\), which is in the range (-1,1). Sufficient.
Can you please explain why, in B above, we must NOT change the sign in |x+1| leaving it as x+1, and must change the sign in 2|x-1| making it 2(-x+1)?
Thank you.
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: \(|5|=5\);
For B. if \(-1\leq{x}\leq{1}\) (green range) --> then \(x+1\geq{0}\) (try some value of \(x\) from the given range to check: for example \(x=0\)) so \(|x+1|=x+1\) BUT \(x-1\leq{0}\) ] (again try some value of \(x\) from the given range to check: for example \(x=0\)) so \(|x-1|=-(x-1)\) thus \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\).
Re: Inequality - absolute on both sides [#permalink]
04 Feb 2011, 11:19
2
This post received KUDOS
The question is
Is |x| < 1 ?
Asking whether x falls between 1 and -1, exclusive.
if |x|<1 x<1; 0.5, 0.2,0 and x>-1; -0.5,-0.2
or -1<x<1
Moment you write x=1 or x=-1; the |x|<1 becomes false.
(1) |x + 1| = 2|x – 1|
For modulus on both sides:
Case I. solve the equation with no sign change
so +(x+1) = +2(x-1) x+1=2x-2 x=3. Not between -1 and 1.
Case II. solve the equation with sign change on one side. Either RHS or LHS. Let's do the sign change on LHS
so -(x+1) = +2(x-1) -x-1=2x-2 3x=1. x=1/3 x is between -1 and 1.
CaseII(b): Even if we did sign change on RHS, we would have gotten the same result. Let's try +(x+1) = -2(x-1) x+1=-2x+2 3x=1 x=1/3. Same result as before.
So; sign change should be done for either LHS or RHS.
Now, we have two solutions for x; 1/3(between -1 and 1), 3(not between -1 and 1)
Not sufficient.
###Also please substitute these factors of x into the main equation and check whether the factors indeed satisfy the equation. because say if 3 didn't satisfy the equation and 1/3 does. The statement would be sufficient.###
(2) |x – 3| > 0
Here modulus only on LHS; So, try this with both signs;
+(x-3) > 0 x-3>0 x>3
and
-(x-3) > 0 -x+3>0 -x>-3 x<3
Here x can be either less than 3 or more than 3. Not 3.
However; this doesn't tell us definitively whether x lies between -1 and 1.
Not Sufficient.
Using both:
We know x can be either (1/3 or 3) by 1st statement. Second statement tells us that x can not be 3.
Re: Inequality - absolute on both sides [#permalink]
04 Feb 2011, 13:47
Expert's post
Matt1177 wrote:
Bunuel wrote:
For B. if \(-1\leq{x}\leq{1}\) (green range) --> then \(x+1\geq{0}\) (try some value of \(x\) from the given range to check: for example \(x=0\)) so \(|x+1|=x+1\) BUT \(x-1\leq{0}\) ] (again try some value of \(x\) from the given range to check: for example \(x=0\)) so \(|x-1|=-(x-1)\) thus \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\).
So, as I understand, we need to plug the value from the available range into the modulus. If the expression in the modulus becomes negative of zero, we must change the sign. And if the expression becomes positive or zero, then we should leave it as it is. Am I correct?
Thanks again for your help, Bunuel.
Yes. You should really try the links I've provided in the above posts.
Matt1177 wrote:
fluke wrote:
Case II. solve the equation with sign change on one side. Either RHS or LHS. Let's do the sign change on LHS
So, we cannot change the signs on both LHS and RHS at once?
No that's not the point. When you have absolute values on both sides of the equation expansion can be either + + (or which is the same - - because \(x + 1 = 2(x - 1)\) is the same as \(-(x + 1) = -2(x - 1)\)) OR + - (or which is the same - + because \(x + 1 = -2(x - 1)\) is the same as \(-(x + 1)=2(x - 1)\)) so basically only two options.
So as I've written in my solution: "alternately you could just expand absolute values of RHS and LHS with same sign (for example both LHS and RHS positive: \(x + 1 = 2(x - 1)\)) and then with different sign (for example LHS positive and RHS negative \(x + 1 = 2(-x + 1)\)), solve for \(x\) both equations, and finally check whether the solutions satisfy \(|x + 1| = 2|x - 1|\)."
Re: Inequality - absolute on both sides [#permalink]
05 Feb 2011, 07:51
Expert's post
ajit257 wrote:
Bunuel...if the second statement was (2) |x – 3| = 0 then in that case could we say
x= 3.
If (2) were |x – 3| = 0 then yes, we would have that x=3 and in this case this statement would be sufficient, as we could answer NO to the question whether -1<x<1. _________________
Mustu, for (1) |x + 1| = 2|x – 1| when looking at X<0 case, how come you multiply only right-hand side by negative, but not both sides? thanks
We have to consider 4 cases totally..
1) + + 2) + - 3) - + 4) - -
If u analyse more closely, then u will find that case 1 and 4 ( - and - cancels out) are the same and cases 2 and 3 are the same( + and - 0r - and + are the same). Hope this is clear.
From Statement 1 square both sides: --> 2^2+2x+1=4x^2-8x+4 --> 0 = 3x^2-10x+3 --> 0=(3x-1)(x-3) x=3 or x=1/3 Hence, insufficient.
From Statement 2 values of x: any number larger or smaller than 3 (i.e. x not equals to 3) Hence, insufficient. From statement 1+2 Sufficient.
Answer: C _________________
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Mustu, for (1) |x + 1| = 2|x – 1| when looking at X<0 case, how come you multiply only right-hand side by negative, but not both sides? thanks
We have to consider 4 cases totally..
1) + + 2) + - 3) - + 4) - -
If u analyse more closely, then u will find that case 1 and 4 ( - and - cancels out) are the same and cases 2 and 3 are the same( + and - 0r - and + are the same). Hope this is clear.
Regards, Mustu
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