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2x+3>x+6 [#permalink]
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29 Dec 2006, 14:39
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2x+3>x+6
please solve



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2x+3>x+6
In order to find the solution by simplifying the inequation (no absolute), we have to work on the following domains:
o x =< 6
o 6 < x < 3/2
o x >= 3/2
o If x =< 6, the inequation is simplified to:
(2*x+3) > (x+6)
<=> x < 3
Thus, all values of x such that x =< 6 are solutions.
o If 6 < x < 3/2, the inequation is simplified to:
(2*x+3) > (x+6)
<=> 3*x < 9
<=> x < 3
Thus, all values of x such that 6 < x < 3 are solutions.
o If x >= 3/2, the inequation is simplified to:
(2*x+3) > (x+6)
<=> x > 3
Thus, all values of x such that x > 3 are solutions.
Finally, we add all solutions :
x < 3 or x > 3



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Fig wrote: 2x+3>x+6
In order to find the solution by simplifying the inequation (no absolute), we have to work on the following domains: o x =< 6 o 6 < x < 3/2 o x >= 3/2
Hi Fig
Can you please explain how do you find the domains.
Thanks



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actually, for this question there is a simpler way to solve....
since both sides of the equation are positive, we can safely square them
(in general, for positive a,b a>b if and only if a^2>b^2)
getting in this case:
4x^2+12x+9 > x^2+12x+36 which simplifies to:
x^2 > 9
which yields easily the result Fig got.



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800_gal wrote: Fig wrote: 2x+3>x+6
In order to find the solution by simplifying the inequation (no absolute), we have to work on the following domains: o x =< 6 o 6 < x < 3/2 o x >= 3/2
Hi Fig Can you please explain how do you find the domains. Thanks
Basically, we look for the "crucial" points that flip the sign in every abs.
Here, we have 2 abs and so 2 point to find by equalling them to 0:
> 2*x+3 = 0 <=> x = 3/2
> x+6 = 0 <=> x = 6
After that, the domain definied from those points.
o x =< 6
o 6 < x < 3/2
o x >= 3/2



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Hi,
I found Hobbit's method of squaring both sides and finding the solution faster and easier. Frankly Fig's method looks a bit daunting. I am still not clear about the concept of domains for absolute values as suggested by Fig.
The way I approached the problem was the traditional way of using all 4 possibilities for setting up absolute values in an inequality.
eg. what I did was I used the following 4 equations.
(1) 2x + 3 > x + 6 yields x > 3
(2) 2x  3 > x + 6 yields x < 3
(3) 2x + 3 > x  6 yields x > 3
(4) 2x  3 > x  6 yields x < 3
Using the more restrictive inequality we arrive at x > 3 or x < 3.
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subhen wrote: Hi,
I found Hobbit's method of squaring both sides and finding the solution faster and easier. Frankly Fig's method looks a bit daunting. I am still not clear about the concept of domains for absolute values as suggested by Fig.
The way I approached the problem was the traditional way of using all 4 possibilities for setting up absolute values in an inequality.
eg. what I did was I used the following 4 equations. (1) 2x + 3 > x + 6 yields x > 3 (2) 2x  3 > x + 6 yields x < 3 (3) 2x + 3 > x  6 yields x > 3 (4) 2x  3 > x  6 yields x < 3
Using the more restrictive inequality we arrive at x > 3 or x < 3.
The way that hobbit uses is good for this question ... Not for all ... And, I do agree that there are different ways to solve the same problem
Personnally, I prefer the graphical approach .... Drawing fastly an XY plan with the courbs and the important points determined by few calculation if required
To come back to your inequations : the (3) and (4) does not represent the original ineqation. There is no reason for the right side to be adapted to x.



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i never tried to claim my method is generic....
in fact it is good only if you know that both sides are positive. this is a special case.
there is also another way that works:
instead of solving something > something else
solve first: something = something else
and then sample (pick numbers) in each region.
for example:
to solve 2x+3>x+6
we start by solving
2x+3=x+6
there are two cases:
2x+3=x+6 => x=3
or
2x+3= (x+6) => x=3
so we have three regions: x<3 , 3<x<3 and x>3
pick number for each region:
try x=4 => we see that 2x+3=5=5 > x+6 = 2 = 2
try x=0 => we see that 2x+3 = 3 < x+6=6
try x=4 => we see that 2x+3 = 11 > x+6=10
so there are two regions that works and this is the solution:
x<3 or x>3
the reason why checking one point of the region is enough to infer that the whole region share the property is beyond the scope of the discussion here (i can tell you if you really want)...
but this is a good method for solving inequalities in almost all cases.



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hobbit wrote: i never tried to claim my method is generic.... in fact it is good only if you know that both sides are positive. this is a special case.
there is also another way that works: instead of solving something > something else solve first: something = something else and then sample (pick numbers) in each region.
for example: to solve 2x+3>x+6 we start by solving 2x+3=x+6 there are two cases: 2x+3=x+6 => x=3 or 2x+3= (x+6) => x=3
so we have three regions: x<3 , 3<x<3 and x>3 pick number for each region: try x=4 => we see that 2x+3=5=5 > x+6 = 2 = 2 try x=0 => we see that 2x+3 = 3 < x+6=6 try x=4 => we see that 2x+3 = 11 > x+6=10
so there are two regions that works and this is the solution: x<3 or x>3
the reason why checking one point of the region is enough to infer that the whole region share the property is beyond the scope of the discussion here (i can tell you if you really want)...
but this is a good method for solving inequalities in almost all cases.
I must disagree here... One more time, it works on specific question. But, what about the others?
Could u define the scope of this method ?... In these cases, people will use this short cut



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Fig wrote: hobbit wrote: the reason why checking one point of the region is enough to infer that the whole region share the property is beyond the scope of the discussion here (i can tell you if you really want)...
but this is a good method for solving inequalities in almost all cases. I must disagree here... One more time, it works on specific question. But, what about the others? Could u define the scope of this method ?... In these cases, people will use this short cut
actually this method works for virtually any inequality !!!!!!
(the real condition for it to work is that both sides will be continuous functions...for those who know the concept.... for practical uses  it is every inequality that is defined for all numbers)
i'll explain more....
the methods work like this:
to solve an inequality: expression1>expression2, where both expressions are defined for all x.
a) solve the equation: expression1=expression2
b) put the solutions on the number line. this divide it into segments.
c) choose one number for each segment created. for each such value  check the original inequality. if it holds  then the whole segment solves the inequality.
the solution is the union of all segments whose representative "passed" the test in stage c.
the reason this method works is this 
suppose i didn't work. this means that there exists two numbers a and b in the same segment for one exp1>exp2 and for the other exp1<exp2.
but (and this is the complicated part of the method). if it was the case, there was a certain number c between a and b for which exp1=exp2. this is the "continuity assumption"  you can't go from negative to positive without crossing through 0.
but there exists such c between a and b  then it means that a and b are NOT in the same segment. which contradicts our assumption.
so there can't be c like that, and it can't be that exp1<exp2 for a and exp1>exp2 for b for every a,b in the same segment.
so it is sufficient to check one number for each segment to check whether the original inequality holds for the segment.
this method can be extended to work for expressions that have some points where they are undefined.....
and to calm all people down.... most if not all expressions that you'll get into in life are continuous. so the method is quite robust.



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Fig wrote: An exemple my saying?... Try this one:
x+2 > x^2.
Following your theory, when x > +1/2, the behavior is all time similar.
So, x = 3/2 >>> 3/2 + 2 = 7/2 (left side) and 9/4 right side.... Thus, the inequation is verified for all x > 1/2 ?
No... If we take x = 4, we have 6 > 16 ?.
And also, both sides have continuity in their fonctions.
ok... lets try solving this in "my method"...
the inequation is x+2>X^2
so first we solve:
x+2=x^2
for x>2 we get: x+2=x^2 => x=2 or x=1
for x<2: x2=x^2 which has no solution.
so our solutions to the equation divide the number line into 3 segments:
x<1
1<x<2
x>2
checking the first (using x=2) we see that the inequation doesn't hold
checking the middle segment (using x=0) we see that the inequation holds
checking the last segment (using x=3) the inequation doesn't hold.
so the solution is the middle segment.
1<x<2
i'm confident that any solving method you use will lead you to this solution.



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hobbit wrote: Fig wrote: An exemple my saying?... Try this one:
x+2 > x^2.
Following your theory, when x > +1/2, the behavior is all time similar.
So, x = 3/2 >>> 3/2 + 2 = 7/2 (left side) and 9/4 right side.... Thus, the inequation is verified for all x > 1/2 ?
No... If we take x = 4, we have 6 > 16 ?.
And also, both sides have continuity in their fonctions. ok... lets try solving this in "my method"... the inequation is x+2>X^2 so first we solve: x+2=x^2 for x>2 we get: x+2=x^2 => x=2 or x=1 for x<2: x2=x^2 which has no solution. so our solutions to the equation divide the number line into 3 segments: x<1 1<x<2 x>2 checking the first (using x=2) we see that the inequation doesn't hold checking the middle segment (using x=0) we see that the inequation holds checking the last segment (using x=3) the inequation doesn't hold. so the solution is the middle segment. 1<x<2 i'm confident that any solving method you use will lead you to this solution.
Sorry hobbit ... I had deleted my post ....
An example should be this one and I follow your way :
x^2  4 > x + 2.
So, x^2  x  6 = 0.... roots = 2 and 3.
That implies, we study on x < 2, 2 < x < 3 and x > 3.
So, in the case of 2 < x < 3, picking 1 number is sufficient. We could take 2.
2^2  4 = 0 < 2 + 2 = 4.
So, the inequality is never verified on 2 < x < 3 ?.... The answer is no. Why?
X = 1 >>> 3 = 3 > 1+2 = 1.



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you just missed one solution for the equation....
x^24=x+2 has 3 solutions: x=3 x=1 and x=2
so there are 4 segments to check  not three.
note that you checked x=2 and tried to give a counter example with x=1. given that x=1 is also a solution for the equation, 2 and 1 are not in the same segment  so no contradiction.
"my method" still works.... (and i'm quite sure it will be very difficult to find a counter example).



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hobbit wrote: you just missed one solution for the equation....
x^24=x+2 has 3 solutions: x=3 x=1 and x=2
so there are 4 segments to check  not three.
note that you checked x=2 and tried to give a counter example with x=1. given that x=1 is also a solution for the equation, 2 and 1 are not in the same segment  so no contradiction.
"my method" still works.... (and i'm quite sure it will be very difficult to find a counter example).
Ahhh... I c. U preserve the absolute to solve... A part, I must say, I didnt see.
Yes.... u are checking the sign (table of sign of the function http://www.gmatclub.com/phpbb/viewtopic.php?t=36468) by empirical values. It works yes.... Agree
So, your methodology is useful for cases that do not contain too much absolute values, that should be the GMAT cases.
In longer cases, x+8  x7 > x+1 + x1 equalling makes it hard to solve.



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Re: inequality + absolute [#permalink]
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31 Dec 2006, 01:57
old_dream_1976 wrote: 2x+3>x+6
please solve
If you feel comfortable plotting lines on the xy plane, you could solve this inequality just by inspecting the intersections of both absolute value functions. The result is
x < 3 AND x > 3.



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Re: inequality + absolute [#permalink]
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12 Aug 2009, 04:07
I didn't any solution. There can be four situations: First, when 2x+3is positive and x+6 is positive => 2x+3 > x+6 => x > 3 Second, when 2x+3is positive and x+6 is negative => 2x+3 > x6 => 3x > 9 => x > 3 Third, when 2x+3is negative and x+6 is positive => 2x3 > x+6 => 3x > 9 => x < 3 Fourth, when 2x+3is negative and x+6 is negative => 2x3 > x6 => 3x > 9 => x < 3 Now, combining all four, we have no possible solution. Can anybody pls explain how to solve this by opening modules?
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Re: inequality + absolute [#permalink]
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16 Dec 2011, 12:38
My answer: \(x > 3\) and \(x < 3\) Previously to that answer, I got the possible solutions: \(x > 3\), \(x < 3\), \(x < 3\), and \(x > 3\). So, I replaced these values in the original inequlity to see whether all these solutions confirmed that relationship. Only, the first and third solution did so. Is my approach correct? Thank you.
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Am unable to understand how the solutions have been combined to get x<3 or x>3.. The solution does not take into account the limits set by 6 in x<6 or 6<x<3.. Please Help Fig wrote: 2x+3>x+6
In order to find the solution by simplifying the inequation (no absolute), we have to work on the following domains: o x =< 6 o 6 < x < 3/2 o x >= 3/2
o If x =< 6, the inequation is simplified to:
(2*x+3) > (x+6) <=> x < 3
Thus, all values of x such that x =< 6 are solutions.
o If 6 < x < 3/2, the inequation is simplified to:
(2*x+3) > (x+6) <=> 3*x < 9 <=> x < 3
Thus, all values of x such that 6 < x < 3 are solutions.
o If x >= 3/2, the inequation is simplified to:
(2*x+3) > (x+6) <=> x > 3
Thus, all values of x such that x > 3 are solutions.
Finally, we add all solutions : x < 3 or x > 3



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Re: inequality + absolute [#permalink]
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14 Feb 2012, 00:27
You can solve this question using Mgmat equalities approach which is very easy.
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devinawilliam83 wrote: Am unable to understand how the solutions have been combined to get x<3 or x>3.. The solution does not take into account the limits set by 6 in x<6 or 6<x<3.. Please Help Approach #1: Fig's approach.\(2x+3>x+6\) We have two check points: \(6\) and \(\frac{3}{2}\) (check point the value fox for which an expression in modulus equals to zero). Thus 3 ranges to check: 1. \(x\leq{6}\) > in this range \(2x+3<0\) and \(x+6<0\) thus absolute values expand as: \((2x+3)>(x+6)\) > \(x<3\). Now, since we consider the range \(x\leq{6}\) then the solution for this range will be \(x<6\): (6)(3/2)2. \(6<x<\frac{3}{2}\) > in this range \(2x+3<0\) and \(x+6>0\) thus absolute values expand as: \((2x+3)>(x+6)\) > \(x<3\). Since we consider the range \(6<x<\frac{3}{2}\) then the solution for this range will be \(6<x<3\): (6)(3)(3/2)3. \(\frac{3}{2}\leq{x}\) > in this range \(2x+3>0\) and \(x+6>0\) thus absolute values expand as: \((2x+3)>(x+6)\) > \(x>3\). Since we consider the range [\(\frac{3}{2}\leq{x}\) then the solution for this range will be \(x>3\): (6)(3)(3/2)3Combined solution is: (6)(3)(3/2)3\(x<3\) or \(x>3\). Approach #2: Easier one.\(2x+3>x+6\) Since both sides of the inequality are nonnegative then we can safely square it: \((2x+3)^2>(x+6)^2\) > \(4x^2+12x+9>x^2+12x+36\) > \(3x^2>27\) > \(x^2>9\) > \(x<3\) or \(x>3\). Hope it helps.
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