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Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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Which values of x are solutions to the inequality x + 1 + x  1 <= 2 ? Apart from algebra can we think conceptually to solve this



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Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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sandal85 wrote: Which values of x are solutions to the inequality x + 1 + x  1 <= 2 ? Apart from algebra can we think conceptually to solve this \(x + 1+x  1\leq{2}\) > we have two check points 1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check. If \(x<1\) then \(x + 1+x  1\leq{2}\) expands as \((x+1)(x1)\leq{2}\) > \(x\geq{1}\), not a valid range since we are considering \(x<1\); If \(1\leq{x}\leq{1}\) then \(x + 1+x  1\leq{2}\) expands as \(x+1(x1)\leq{2}\) > \(2\leq{2}\) which is true, so for \(1\leq{x}\leq{1}\) given inequality holds true; If \(x>1\) then \(x + 1+x  1\leq{2}\) expands as \(x+1+x1\leq{2}\) > \(x\leq{1}\), not a valid range since we are considering \(x>1\). So, finally we have that \(x + 1+x  1\leq{2}\) holds true for \(1\leq{x}\leq{1}\). For more check Absolute Value chapter of Math Book: http://gmatclub.com/forum/mathabsolute ... 86462.htmlHope it helps.
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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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22 May 2012, 01:28
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Bunuel wrote: sandal85 wrote: Which values of x are solutions to the inequality x + 1 + x  1 <= 2 ? Apart from algebra can we think conceptually to solve this \(x + 1+x  1\leq{2}\) > we have two check points 1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check. If \(x<1\) then \(x + 1+x  1\leq{2}\) expands as \((x+1)(x1)\leq{2}\) > \(x\geq{1}\), not a valid range since we are considering \(x<1\); If \(1\leq{x}\leq{1}\) then \(x + 1+x  1\leq{2}\) expands as \(x+1(x1)\leq{2}\) > \(0\leq{2}\) which is true, so for \(1\leq{x}\leq{1}\) given inequality holds true; If \(x>1\) then \(x + 1+x  1\leq{2}\) expands as \(x+1+x1\leq{2}\) > \(x\leq{1}\), not a valid range since we are considering \(x>1\). So, finally we have that \(x + 1+x  1\leq{2}\) holds true for \(1\leq{x}\leq{1}\). For more check Absolute Value chapter of Math Book: mathabsolutevaluemodulus86462.htmlHope it helps. Dear Bunuel, what these kind of inequality questions are asking for? is it the values of X, for which the solution of equation is <=2? i had been through the link provided and have the same doubt regarding the example 1  x+34x=8+x How many solutions does the equation have? How you have taken "four conditions given"?



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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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22 May 2012, 10:25
kashishh wrote: Bunuel wrote: sandal85 wrote: Which values of x are solutions to the inequality x + 1 + x  1 <= 2 ? Apart from algebra can we think conceptually to solve this \(x + 1+x  1\leq{2}\) > we have two check points 1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check. If \(x<1\) then \(x + 1+x  1\leq{2}\) expands as \((x+1)(x1)\leq{2}\) > \(x\geq{1}\), not a valid range since we are considering \(x<1\); If \(1\leq{x}\leq{1}\) then \(x + 1+x  1\leq{2}\) expands as \(x+1(x1)\leq{2}\) > \(0\leq{2}\) which is true, so for \(1\leq{x}\leq{1}\) given inequality holds true; If \(x>1\) then \(x + 1+x  1\leq{2}\) expands as \(x+1+x1\leq{2}\) > \(x\leq{1}\), not a valid range since we are considering \(x>1\). So, finally we have that \(x + 1+x  1\leq{2}\) holds true for \(1\leq{x}\leq{1}\). For more check Absolute Value chapter of Math Book: mathabsolutevaluemodulus86462.htmlHope it helps. Dear Bunuel, what these kind of inequality questions are asking for? is it the values of X, for which the solution of equation is <=2? i had been through the link provided and have the same doubt regarding the example 1  x+34x=8+x How many solutions does the equation have? How you have taken "four conditions given"? The question asks about the range(s) of x for which the given inequality holds true. We have that it holds for \(1\leq{x}\leq{1}\), for example if x=0 then \(x + 1+x  1\leq{2}\) is true but if x is out of this range, say x=3, then \(x + 1+x  1\leq{2}\) is NOT true. Hope it's clear.
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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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23 May 2012, 01:45
Thankyou Bunuel, It is asking for range. i understand,so is in this question "x+34x=8+x How many solutions does the equation have?
Please can you elaborate on "four conditions given"?



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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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23 May 2012, 06:05
If for the same question, x+34x=8+x, we had to identify for what what value of x will the equation be valid. How would we do that?



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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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23 May 2012, 06:29
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kashishh wrote: Thankyou Bunuel, It is asking for range. i understand,so is in this question "x+34x=8+x How many solutions does the equation have?
Please can you elaborate on "four conditions given"? pavanpuneet wrote: If for the same question, x+34x=8+x, we had to identify for what what value of x will the equation be valid. How would we do that? How many solutions does \(x+3  4x = 8+x\) have? Basically the same here: we have three check points 8, 3 and 4 and thus four ranges to check. If \(x < 8\) then \(x+3  4x = 8+x\) expands as \((x+3)(4x)=(8+x)\) > \(x = 1\), which is not a valid solution since we are considering \(x < 8\) range and 1 is out of it; If \(8\leq{x}\leq{3}\) then \(x+3  4x = 8+x\) expands as \((x+3)(4x)=(8+x)\) > \(x=15\), which is also not a valid solution since we are considering \(8\leq{x}\leq{3}\) range; If \(3<x<4\) then \(x+3  4x = 8+x\) expands as \((x+3)(4x)=(8+x)\) > \(x = 9\), which is also not a valid solution since we are considering \(3<x<4\) range; If \(x\geq{4}\) then \(x+34x=8+x\) expands as \((x+3)+(4x)=(8+x)\) > \(x = 1\), which is also not a valid solution since we are considering \(x>4\) range. So we have that no solution exist for \(x+34x=8+x\) (no x can satisfy this equation). Hope it's clear.
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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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23 May 2012, 07:29
Inequalities have been quite a nightmare for me.. I understood the overall solution but, based on what do we decide to x<8 and not x<=8....?? I know this may be a very basic question, but can you please explain this concept for these ranges. Will be a great help! Thanks in advance.



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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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23 May 2012, 21:03
Hi, Bunuel! How many solutions does have?
In your solution when you define intervals, how do you define what inequality sign to put? I mean < or> or<= For example, why the second interval is 8<=x<=3, why both inequality signs are <=?
Thank you in advance



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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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24 May 2012, 00:51
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pavanpuneet wrote: Inequalities have been quite a nightmare for me.. I understood the overall solution but, based on what do we decide to x<8 and not x<=8....?? I know this may be a very basic question, but can you please explain this concept for these ranges. Will be a great help! Thanks in advance. Rice wrote: Hi, Bunuel! How many solutions does have?
In your solution when you define intervals, how do you define what inequality sign to put? I mean < or> or<= For example, why the second interval is 8<=x<=3, why both inequality signs are <=?
Thank you in advance We should include "check points" (8, 3, and 4) in either of the ranges but it really doesn't matter in which one. So, for example we could have written: If \(x\leq{8}\) ... If \(8<x<3\) ... If \(3\leq{x}<\leq{4}\) ... If \(x>4\) ... Hope it's clear.
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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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24 May 2012, 02:52
Bunuel wrote: How many solutions does \(x+3  4x = 8+x\) have?
Basically the same here: we have three check points 8, 3 and 4 and thus four ranges to check.
If \(x < 8\) then \(x+3  4x = 8+x\) expands as \((x+3)(4x)=(8+x)\) > \(x = 1\), which is not a valid solution since we are considering \(x < 8\) range and 1 is out of it;
If \(8\leq{x}\leq{3}\) then \(x+3  4x = 8+x\) expands as \((x+3)(4x)=(8+x)\) > \(x=15\), which is also not a valid solution since we are considering \(8\leq{x}\leq{3}\) range;
If \(3<x<4\) then \(x+3  4x = 8+x\) expands as \((x+3)(4x)=(8+x)\) > \(x = 9\), which is also not a valid solution since we are considering \(3<x<4\) range;
If \(x\geq{4}\) then \(x+34x=8+x\) expands as \((x+3)+(4x)=(8+x)\) > \(x = 1\), which is also not a valid solution since we are considering \(x>4\) range.
So we have that no solution exist for \(x+34x=8+x\) (no x can satisfy this equation).
Hope it's clear. The implicit understanding is that when a equation for a range is simplified and they contradict the basic premises of the considered range then the range is not considered as the solution.
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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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24 Nov 2012, 00:18
Hi Bunnel,
I follow all your solutions. Really great!!!. Thanks for taking time to answer us. I read the anwer explanations for the ques above. But still I am little unclear. Please help me.
I know x1 = 4 can be written as (x1) = 4 and (x1) = 4. Fine.
But for the Ques x+3  4x = x+8 I understand the key points are 8, 3, 4
For x<=  8, First how did you take the condition (x+3)(4x)=(8+x) ? Why not (x+3)(4x) = (x+8) or (x+3)+(4x) = (8+x)?
And why x <= 8 is tested as condition? why not x >= 8.
Regards, Sree



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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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14 May 2015, 12:00
Bunuel wrote: sandal85 wrote: Which values of x are solutions to the inequality x + 1 + x  1 <= 2 ? Apart from algebra can we think conceptually to solve this \(x + 1+x  1\leq{2}\) > we have two check points 1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check. If \(x<1\) then \(x + 1+x  1\leq{2}\) expands as \((x+1)(x1)\leq{2}\) > \(x\geq{1}\), not a valid range since we are considering \(x<1\); If \(1\leq{x}\leq{1}\) then \(x + 1+x  1\leq{2}\) expands as \(x+1(x1)\leq{2}\) > \(0\leq{2}\) which is true, so for \(1\leq{x}\leq{1}\) given inequality holds true; If \(x>1\) then \(x + 1+x  1\leq{2}\) expands as \(x+1+x1\leq{2}\) > \(x\leq{1}\), not a valid range since we are considering \(x>1\). So, finally we have that \(x + 1+x  1\leq{2}\) holds true for \(1\leq{x}\leq{1}\). For more check Absolute Value chapter of Math Book: mathabsolutevaluemodulus86462.htmlHope it helps. Hi Bunuel, Should the ranges for checking be: a) x < 1 b) 1 =< x < 1 c) x >= 1 ? Have you consciously altered the position of the greater than or equal and less than or equal signs for the three ranges because it would make no difference? Hope my question is clear. TIA.



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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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14 May 2015, 23:38
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sandal85 wrote: Which values of x are solutions to the inequality x + 1 + x  1 <= 2 ? Apart from algebra can we think conceptually to solve this You can solve this question using the number line approach too. "Distance of x from 1" + "Distance of x from 1" <= 2 (2)(1)0(1)(2) Note that distance between 1 and 1 is 2 so whenever x is between these two values, the sum of distance from 1 and 1 will be 2 (2)(1)x0(1)(2) Hence all values between 1 and 1 (inclusive) will satisfy this condition. When you go to the right of 1 or left of 1, the sum of distances from 1 and 1 will exceed 2. So the only range that satisfies the inequality is 1 <= x <= 1
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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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11 Sep 2015, 23:34
Bunuel wrote: sandal85 wrote: Which values of x are solutions to the inequality x + 1 + x  1 <= 2 ? Apart from algebra can we think conceptually to solve this \(x + 1+x  1\leq{2}\) > we have two check points 1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check. If \(x<1\) then \(x + 1+x  1\leq{2}\) expands as \((x+1)(x1)\leq{2}\) > \(x\geq{1}\), not a valid range since we are considering \(x<1\); If \(1\leq{x}\leq{1}\) then \(x + 1+x  1\leq{2}\) expands as \(x+1(x1)\leq{2}\) > \(0\leq{2}\) which is true, so for \(1\leq{x}\leq{1}\) given inequality holds true;If \(x>1\) then \(x + 1+x  1\leq{2}\) expands as \(x+1+x1\leq{2}\) > \(x\leq{1}\), not a valid range since we are considering \(x>1\). So, finally we have that \(x + 1+x  1\leq{2}\) holds true for \(1\leq{x}\leq{1}\). For more check Absolute Value chapter of Math Book: mathabsolutevaluemodulus86462.htmlHope it helps. Dear BunuelI didnt get one point in highlighted part: x + 1  (x  1) <= 2 => x + 1  x + 1 <=2 => 2 <= 2 But you are getting 0 <=2 Did i miss anything?
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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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19 Oct 2015, 03:09
Bunuel wrote: sandal85 wrote: Which values of x are solutions to the inequality x + 1 + x  1 <= 2 ? Apart from algebra can we think conceptually to solve this \(x + 1+x  1\leq{2}\) > we have two check points 1 and 1 (check point is the value of x for which an absolute value equals to zero) and thus three ranges to check. If \(x<1\) then \(x + 1+x  1\leq{2}\) expands as \((x+1)(x1)\leq{2}\) > \(x\geq{1}\), not a valid range since we are considering \(x<1\); If \(1\leq{x}\leq{1}\) then \(x + 1+x  1\leq{2}\) expands as \(x+1(x1)\leq{2}\) > \(0\leq{2}\) which is true, so for \(1\leq{x}\leq{1}\) given inequality holds true; If \(x>1\) then \(x + 1+x  1\leq{2}\) expands as \(x+1+x1\leq{2}\) > \(x\leq{1}\), not a valid range since we are considering \(x>1\). So, finally we have that \(x + 1+x  1\leq{2}\) holds true for \(1\leq{x}\leq{1}\). For more check Absolute Value chapter of Math Book: mathabsolutevaluemodulus86462.htmlHope it helps. Hi Bunuel, I have one questions regarding absolute value equations: why can't we use the same logis (as stated below) to solve this kind of equations > x + 1 + x  1 <= 2. We check here ranges, but in the example below we don't need doing so. Thanks in advance. 2x  1 = 4x + 9 Solution: x =4/3 or5 2x1=4x+9 or 2x1=(4x+9)
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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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19 Oct 2015, 19:04
Bunuel wrote: If \(1\leq{x}\leq{1}\) then \(x + 1+x  1\leq{2}\) expands as \(x+1(x1)\leq{2}\) > \(0\leq{2}\) which is true, so for \(1\leq{x}\leq{1}\) given inequality holds true;
Bunuel, there is a slight typo. It should read \(2\leq{2}\) instead of \(0\leq{2}\)



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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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12 Nov 2015, 21:06
Bunuel wrote: pavanpuneet wrote: Inequalities have been quite a nightmare for me.. I understood the overall solution but, based on what do we decide to x<8 and not x<=8....?? I know this may be a very basic question, but can you please explain this concept for these ranges. Will be a great help! Thanks in advance. Rice wrote: Hi, Bunuel! How many solutions does have?
In your solution when you define intervals, how do you define what inequality sign to put? I mean < or> or<= For example, why the second interval is 8<=x<=3, why both inequality signs are <=?
Thank you in advance We should include "check points" (8, 3, and 4) in either of the ranges but it really doesn't matter in which one. So, for example we could have written: If \(x\leq{8}\) ... If \(8<x<3\) ... If \(3\leq{x}<\leq{4}\) ... If \(x>4\) ... Hope it's clear. Hi Bunuel, For the question posted, if the ranges where x<= 1, then (x+1)(x1) <=2, gives x>=1 (Not Satisfied) 1<x<1, then (x+1) (x1) <=2 , gives 2 <=2 (Satisfied) x>=1, then (x+1)+(x1)<=2, gives x<=1(Not Satisfied) So I will choose the range as 1<x<1. But the correct answer is 1<=x<=1. What am i missing? Regards, Deepthi



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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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13 May 2016, 00:12
Hi,
is it correct if i follow below approach to solve this
x+1 + x1 <=2
2x <=2 > this expands to
2x <=2 or 2x <=2
x<=1 or x>=1 i.e. 1<=x<=1



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Re: Which values of x are solutions x + 1 + x  1 <= 2 [#permalink]
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15 May 2016, 03:15
sandal85 wrote: Which values of x are solutions to the inequality x + 1 + x  1 <= 2 ? Apart from algebra can we think conceptually to solve this We need to find the value of x such that the sum of the distance of x from 1 and 1 is less than or equal to 2. If the value of x is taken less than 1, then the distance of x from 1 will be more than 2. Similarly, if value of x is greater than 1, then the distance of x from 1 will be greater than 2. So, the only possible region where the sum can be <=2 is 1<=x<=1




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