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25 Sep 2012, 09:21
Kudos
Bookmarks
Hi Experts,
I need help with inequalities.
I am finding difficulty in finding solutions to inequalities in complex problems. I understood the concept of finding the solution in simpler problems (to the extent of finding the range and plotting it on the number line). But in the below example, I understand till identifying the check points. Also understood the point of using the check points to create ranges for verification. But could not understand how in each of the scenarios the equation is expanded. On what grounds -ve sings are applied to the equations while expanding.
I have gone through MGMAT material on Inequalities. But could not relate this approach anything there. Also checked the GMAT club math book example. Guess even that is on the same lines and could not understand it for the same reasons.
Could someone please explain the strategy/concept here. What basics am I missing here. Are there any links or material I can refer to better understand the basics I am missing. Help on this will be greatly appreciated.
How many solutions does \(|x+3| - |4-x| = |8+x|\) have?
Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.
If \(x < -8\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=-(8+x)\) --> \(x = -1\), 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 \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=(8+x)\) --> \(x=-15\), which is also not a valid solution since we are considering \(-8\leq{x}\leq{-3}\) range;
If \(-3<x<4\) then \(|x+3| - |4-x| = |8+x|\) expands as \((x+3)-(4-x)=(8+x)\) --> \(x = 9\), which is also not a valid solution since we are considering \(-3<x<4\) range;
If \(x\geq{4}\) then \(|x+3|-|4-x|=|8+x|\) expands as \((x+3)+(4-x)=(8+x)\) --> \(x = -1\), which is also not a valid solution since we are considering \(x>4\) range.
So we have that no solution exist for \(|x+3|-|4-x|=|8+x|\) (no x can satisfy this equation).
Thanks in anticipation.
I need help with inequalities.
I am finding difficulty in finding solutions to inequalities in complex problems. I understood the concept of finding the solution in simpler problems (to the extent of finding the range and plotting it on the number line). But in the below example, I understand till identifying the check points. Also understood the point of using the check points to create ranges for verification. But could not understand how in each of the scenarios the equation is expanded. On what grounds -ve sings are applied to the equations while expanding.
I have gone through MGMAT material on Inequalities. But could not relate this approach anything there. Also checked the GMAT club math book example. Guess even that is on the same lines and could not understand it for the same reasons.
Could someone please explain the strategy/concept here. What basics am I missing here. Are there any links or material I can refer to better understand the basics I am missing. Help on this will be greatly appreciated.
How many solutions does \(|x+3| - |4-x| = |8+x|\) have?
Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.
If \(x < -8\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=-(8+x)\) --> \(x = -1\), 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 \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=(8+x)\) --> \(x=-15\), which is also not a valid solution since we are considering \(-8\leq{x}\leq{-3}\) range;
If \(-3<x<4\) then \(|x+3| - |4-x| = |8+x|\) expands as \((x+3)-(4-x)=(8+x)\) --> \(x = 9\), which is also not a valid solution since we are considering \(-3<x<4\) range;
If \(x\geq{4}\) then \(|x+3|-|4-x|=|8+x|\) expands as \((x+3)+(4-x)=(8+x)\) --> \(x = -1\), which is also not a valid solution since we are considering \(x>4\) range.
So we have that no solution exist for \(|x+3|-|4-x|=|8+x|\) (no x can satisfy this equation).
Thanks in anticipation.
25 Sep 2012, 11:39
Kudos
Bookmarks
GetThisDone
Hi Experts,
I need help with inequalities.
I am finding difficulty in finding solutions to inequalities in complex problems. I understood the concept of finding the solution in simpler problems (to the extent of finding the range and plotting it on the number line). But in the below example, I understand till identifying the check points. Also understood the point of using the check points to create ranges for verification. But could not understand how in each of the scenarios the equation is expanded. On what grounds -ve sings are applied to the equations while expanding.
I have gone through MGMAT material on Inequalities. But could not relate this approach anything there. Also checked the GMAT club math book example. Guess even that is on the same lines and could not understand it for the same reasons.
Could someone please explain the strategy/concept here. What basics am I missing here. Are there any links or material I can refer to better understand the basics I am missing. Help on this will be greatly appreciated.
How many solutions does \(|x+3| - |4-x| = |8+x|\) have?
Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.
If \(x < -8\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=-(8+x)\) --> \(x = -1\), 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 \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=(8+x)\) --> \(x=-15\), which is also not a valid solution since we are considering \(-8\leq{x}\leq{-3}\) range;
If \(-3<x<4\) then \(|x+3| - |4-x| = |8+x|\) expands as \((x+3)-(4-x)=(8+x)\) --> \(x = 9\), which is also not a valid solution since we are considering \(-3<x<4\) range;
If \(x\geq{4}\) then \(|x+3|-|4-x|=|8+x|\) expands as \((x+3)+(4-x)=(8+x)\) --> \(x = -1\), which is also not a valid solution since we are considering \(x>4\) range.
So we have that no solution exist for \(|x+3|-|4-x|=|8+x|\) (no x can satisfy this equation).
Thanks in anticipation.
I need help with inequalities.
I am finding difficulty in finding solutions to inequalities in complex problems. I understood the concept of finding the solution in simpler problems (to the extent of finding the range and plotting it on the number line). But in the below example, I understand till identifying the check points. Also understood the point of using the check points to create ranges for verification. But could not understand how in each of the scenarios the equation is expanded. On what grounds -ve sings are applied to the equations while expanding.
I have gone through MGMAT material on Inequalities. But could not relate this approach anything there. Also checked the GMAT club math book example. Guess even that is on the same lines and could not understand it for the same reasons.
Could someone please explain the strategy/concept here. What basics am I missing here. Are there any links or material I can refer to better understand the basics I am missing. Help on this will be greatly appreciated.
How many solutions does \(|x+3| - |4-x| = |8+x|\) have?
Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.
If \(x < -8\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=-(8+x)\) --> \(x = -1\), 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 \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=(8+x)\) --> \(x=-15\), which is also not a valid solution since we are considering \(-8\leq{x}\leq{-3}\) range;
If \(-3<x<4\) then \(|x+3| - |4-x| = |8+x|\) expands as \((x+3)-(4-x)=(8+x)\) --> \(x = 9\), which is also not a valid solution since we are considering \(-3<x<4\) range;
If \(x\geq{4}\) then \(|x+3|-|4-x|=|8+x|\) expands as \((x+3)+(4-x)=(8+x)\) --> \(x = -1\), which is also not a valid solution since we are considering \(x>4\) range.
So we have that no solution exist for \(|x+3|-|4-x|=|8+x|\) (no x can satisfy this equation).
Thanks in anticipation.
Use the definition of the absolute value: \(|x| = x\) if \(x\geq0\) and \(|x|=-x\)if \(x<0.\)
Obviously, instead of \(x\) you can have any algebraic expression. So, for example \(|x+8|=x+8\) if \(x+8\geq0\) or \(x\geq-8\) and \(|x+8| = -(x+8)\) if \(x+8<0\) or \(x<-8\), etc.
Absolute value expresses a distance between two numbers and it is always non-negative.
\(|x|\) is the distance on the number line between \(x\) and \(0\). The distance between \(x\) and \(0\) is the same as the distance between \(-x\) and \(0.\)
\(|7|=7\) and \(|-7|=7\). How can we connect \(-7\) and \(7\)? There is no mathematical operation "just drop the minus sign..." But obviously \(7=(-1)*(-7)\).
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Expert reply
25 Sep 2012, 12:11
Kudos
Bookmarks
|x+3| - |4-x| = |8+x|
Hey
I would be happy to help though I did not get your question very clearly. Do you want to know how did the solution find out the ranges as given? Then I will try to help you out.
I assume you know the basic of the integer function which is:
|x|=x when x>=0
|x|=-x when x<0
So, you can see the checkpoints here are -8, -3 and 4
I hope till now you have followed.
Now lets look at each check points and see how the integer function behaves
When \(x<-8,\)
x+3 is negative, so \(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is negative, so \(|8+x|=-(8+x)\)
So the equation, in this case, becomes,
\(-(x+3)-(4-x)=-(8+x)\)
\(=> x=-1\) which violates the range we have considered
Similarly when\(-8<x<-3\)
x+3 is negative, so\(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so\(|8+x|=(8+x)\)
So equation becomes
\(-(x+3)-(4-x)=(8+x)\)
=>\(x=-15\) which violates the range we have considered
Again, when \(-3<x<4\)
x+3 is positive, so \(|x+3|=(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so \(|8+x|=(8+x)\)
So the equation becomes,
\((x+3)-(4-x)=(8+x)\)
\(2x-1=8+x\)
\(X=9\) which violates the range we have considered
Last, when \(x>4\)
x+3 is positive, so \(|x+3|=(x+3)\)
4-x is negative, so \(|4-x|=-(4-x)\)
8+x is positive, so \(|8+x|=(8+x)\)
\((x+3)+(4-x)=(8+x)\)
\(=>x=-1\) which again violates the range
SO no solution
Did this help?
Hey
I would be happy to help though I did not get your question very clearly. Do you want to know how did the solution find out the ranges as given? Then I will try to help you out.
I assume you know the basic of the integer function which is:
|x|=x when x>=0
|x|=-x when x<0
So, you can see the checkpoints here are -8, -3 and 4
I hope till now you have followed.
Now lets look at each check points and see how the integer function behaves
When \(x<-8,\)
x+3 is negative, so \(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is negative, so \(|8+x|=-(8+x)\)
So the equation, in this case, becomes,
\(-(x+3)-(4-x)=-(8+x)\)
\(=> x=-1\) which violates the range we have considered
Similarly when\(-8<x<-3\)
x+3 is negative, so\(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so\(|8+x|=(8+x)\)
So equation becomes
\(-(x+3)-(4-x)=(8+x)\)
=>\(x=-15\) which violates the range we have considered
Again, when \(-3<x<4\)
x+3 is positive, so \(|x+3|=(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so \(|8+x|=(8+x)\)
So the equation becomes,
\((x+3)-(4-x)=(8+x)\)
\(2x-1=8+x\)
\(X=9\) which violates the range we have considered
Last, when \(x>4\)
x+3 is positive, so \(|x+3|=(x+3)\)
4-x is negative, so \(|4-x|=-(4-x)\)
8+x is positive, so \(|8+x|=(8+x)\)
\((x+3)+(4-x)=(8+x)\)
\(=>x=-1\) which again violates the range
SO no solution
Did this help?
. If you could help me with the below, it will help me put this topic to rest. 
