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tzolkin
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How many solutions does this equation have : Eq = + + - 1) 21 May 2003, 10:12
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How many solutions does this equation have :

Eq = [x+1]+[x-1]+[x+5]-[x-5]


1) Eq = 0
2) X is an Integer



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How many solutions does this equation have : Eq = + + - 1) 21 May 2003, 11:37
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A is an answer.

the equation has only one variable. knowing that Eq=0 we will be able to solve it, and hence fin out the number of solutions.

knowing that X is an integer doesn't help. there could be as many answers as there integers. (well, in this case a little less)
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How many solutions does this equation have : Eq = + + - 1) 21 May 2003, 22:25
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I think A is not clearly right

(1) Eq=0, when we solve the equation, we will get that [-5...-1] is the solution. In other words the INFINITE number of solutions. Is it sufficient to answer the initial question? I don't know

(2) Alone is of no use since we don't know Eq.

Combine: 5 solutions {-5 -4 -3 -2 -1} It is more appropriate I think.

Comments?
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How many solutions does this equation have : Eq = + + - 1) 22 May 2003, 05:38
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Nevertheless, the problem is pretty ambiguous. An infinity is an answer for how many. :writer
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How many solutions does this equation have : Eq = + + - 1) 28 May 2003, 19:51
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hi!
my name is noel. i'm a gmat teacher from the Philippines.
i'd like to help evaluate this problem :
--------------------------------------------------------------------------------

How many solutions does this equation have :

Eq = [x+1]+[x-1]+[x+5]-[x-5]


1) Eq = 0
2) X is an Integer


--------------------------------------------------------------------------------


even before you go to the datasets, you can already simplify the equation :
Eq = x + 1 + x - 1 + x + 5 - x + 5
Eq = 2x + 10


dataset1 states that Eq = 0 :
0 = 2x + 10
-5 = x
x has 1 solution.
sufficient, right ?


dataset2 is insufficient because there can be as many soltions as there are integers, as pointed out by rommer.



noel
gmatcrew@yahoo.com
https://www.geocities.com/gmatcrew/halfwits.htm
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How many solutions does this equation have : Eq = + + - 1) 28 May 2003, 22:25
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gmatcrew
hi!
my name is noel. i'm a gmat teacher from the Philippines.
i'd like to help evaluate this problem :
--------------------------------------------------------------------------------

How many solutions does this equation have :

Eq = [x+1]+[x-1]+[x+5]-[x-5]


1) Eq = 0
2) X is an Integer


--------------------------------------------------------------------------------


even before you go to the datasets, you can already simplify the equation :
Eq = x + 1 + x - 1 + x + 5 - x + 5
Eq = 2x + 10


dataset1 states that Eq = 0 :
0 = 2x + 10
-5 = x
x has 1 solution.
sufficient, right ?


dataset2 is insufficient because there can be as many soltions as there are integers, as pointed out by rommer.



noel
gmatcrew@yahoo.com
https://www.geocities.com/gmatcrew/halfwits.htm
Show more



IMHO: Your solution is interesting--simple removing modulus.
You have lost a lot of roots for the initial equation by simply eliminating the signs of modulus. In fact, the initial equation has more than one root--all numbers from [-5...-1] are roots. Even so, I think that it is OK to answer the question. How many solutions? An infinite number...
A should be an answer at least from a math point of view.

However, C makes the answer perfect.
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How many solutions does this equation have : Eq = + + - 1) 29 May 2003, 05:58
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am i missing anything here ?

Eq = [x+1]+[x-1]+[x+5]-[x-5]


i do not see "modulus" here, unless the symbols [ and ] have mathematical meanings other than grouping .

if so, then i was mistaken.


noel
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How many solutions does this equation have : Eq = + + - 1) 29 May 2003, 06:10
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gmatcrew
am i missing anything here ?

Eq = [x+1]+[x-1]+[x+5]-[x-5]


i do not see "modulus" here, unless the symbols [ and ] have mathematical meanings other than grouping .

if so, then i was mistaken.


noel
Show more


simple brakets are (...) -- it would be too easy to discuss the initial case with simple parentheses. The forum is for difficult stuff. :wall
[...] is an absolute value
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How many solutions does this equation have : Eq = + + - 1) 29 May 2003, 06:15
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hi!

i went to the other posts on this site, and it seems that [ and ] are used here as the absolute value symbols.

i hope all of us can just use | and | to make the questions more math-like.

sorry for misinterpreting the initial post.



sincerely,
noel


ps :

How many solutions does this equation have :

Eq = |x+1| + |x-1| + |x+5| - |x-5| ?


1) Eq = 0
2) X is an Integer



is this really it ?
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How many solutions does this equation have : Eq = + + - 1) 29 May 2003, 06:25
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gmatcrew
hi!

i went to the other posts on this site, and it seems that [ and ] are used here as the absolute value symbols.

i hope all of us can just use | and | to make the questions more math-like.

sorry for misinterpreting the initial post.



sincerely,
noel



ps :

How many solutions does this equation have :

Eq = |x+1| + |x-1| + |x+5| - |x-5| ?


1) Eq = 0
2) X is an Integer



is this really it ?
Show more


how to get "|" by using a usual keyboard?
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How many solutions does this equation have : Eq = + + - 1) 29 May 2003, 08:15
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stolyar, you are absolutely right.

I was too lazy to solve the equation. It is a good lesson for me, in the future I won't be giving answers in a rush.

as for "|" sign. one can use "shift", "\" key. I think any keyboard has this key.

regards
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How many solutions does this equation have : Eq = + + - 1) 29 May 2003, 22:15
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rommer
stolyar, you are absolutely right.

I was too lazy to solve the equation. It is a good lesson for me, in the future I won't be giving answers in a rush.

as for "|" sign. one can use "shift", "" key. I think any keyboard has this key.

regards
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let me try "shift"+"" |||||| amazing... thank you. I have tried to use this symbol for months but have not known how. Now I know. Hurray!!! :wave
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How many solutions does this equation have : Eq = + + - 1) 10 Feb 2004, 16:38
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Can somebody explain how stolyar got this ?

Quote:
Eq=0, when we solve the equation, we will get that [-5...-1] is the solution. In other words the INFINITE number of solutions. Is it sufficient to answer the initial question? I don't know
Show more


How the solution set is arrived at as [-5......-1]?
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How many solutions does this equation have : Eq = + + - 1) 14 Feb 2004, 02:58
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gmatblast
Can somebody explain how stolyar got this ?

Quote:
Eq=0, when we solve the equation, we will get that [-5...-1] is the solution. In other words the INFINITE number of solutions. Is it sufficient to answer the initial question? I don't know

How the solution set is arrived at as [-5......-1]?
Show more


One way to solve this type of problem is to solve the problem in "each interval of interest" -- i.e., an interval where the sign of an expression will not change inside the | | signs. This way, you can solve the problem by taking out the || sign for that interval.

The original expression is:

|x+1|+|x-1|+|x+5|-|x-5|

and we set it equal to zero.

Let's look at the interval between -1 and -5. If -1<x<-5 then
|x+1| is always negative so we can replace it with -(x+1);
|x-1| is always negative so we can replace it with (x-1)
|x+5| is always positive so we can replace it with (x+5)
|x-5| is always negative so we can replace it with -(x+5).

Now we have:
(x+1)-(x+1)+(x+5)+(x-5) = 0 or
0 = 0

which means that in the interval -1<x<5, x can take on any value.

If you try to solve for any other interval you will not be able to get any solutions. Hence, the answer is: [-5,-1] (i.e.; the interval between -5 and -1 inclusive) which has an infinite number of values in the real number domain.
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How many solutions does this equation have : Eq = + + - 1) 29 Dec 2004, 14:30
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Excellant and Simple way of solving the problem.Thank you for posting the answer.Rgds,

Anna

AkamaiBrah
gmatblast
Can somebody explain how stolyar got this ?

Quote:
Eq=0, when we solve the equation, we will get that [-5...-1] is the solution. In other words the INFINITE number of solutions. Is it sufficient to answer the initial question? I don't know

How the solution set is arrived at as [-5......-1]?

One way to solve this type of problem is to solve the problem in "each interval of interest" -- i.e., an interval where the sign of an expression will not change inside the | | signs. This way, you can solve the problem by taking out the || sign for that interval.

The original expression is:

|x+1|+|x-1|+|x+5|-|x-5|

and we set it equal to zero.

Let's look at the interval between -1 and -5. If -1<x<-5 then
|x+1| is always negative so we can replace it with -(x+1);
|x-1| is always negative so we can replace it with (x-1)
|x+5| is always positive so we can replace it with (x+5)
|x-5| is always negative so we can replace it with -(x+5).

Now we have:
(x+1)-(x+1)+(x+5)+(x-5) = 0 or
0 = 0

which means that in the interval -1<x<5, x can take on any value.

If you try to solve for any other interval you will not be able to get any solutions. Hence, the answer is: [-5,-1] (i.e.; the interval between -5 and -1 inclusive) which has an infinite number of values in the real number domain.
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How many solutions does this equation have : Eq = + + - 1) 31 Dec 2004, 05:53
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gmatblast
Can somebody explain how stolyar got this ?

Quote:
Eq=0, when we solve the equation, we will get that [-5...-1] is the solution. In other words the INFINITE number of solutions. Is it sufficient to answer the initial question? I don't know

How the solution set is arrived at as [-5......-1]?

One way to solve this type of problem is to solve the problem in "each interval of interest" -- i.e., an interval where the sign of an expression will not change inside the | | signs. This way, you can solve the problem by taking out the || sign for that interval.

The original expression is:

|x+1|+|x-1|+|x+5|-|x-5|

and we set it equal to zero.

Let's look at the interval between -1 and -5. If -1<x<-5 then
|x+1| is always negative so we can replace it with -(x+1);
|x-1| is always negative so we can replace it with (x-1)
|x+5| is always positive so we can replace it with (x+5)
|x-5| is always negative so we can replace it with -(x+5).

Now we have:
(x+1)-(x+1)+(x+5)+(x-5) = 0 or
0 = 0

which means that in the interval -1<x<5, x can take on any value.

If you try to solve for any other interval you will not be able to get any solutions. Hence, the answer is: [-5,-1] (i.e.; the interval between -5 and -1 inclusive) which has an infinite number of values in the real number domain.
Show more


Thanks for the posts. However, I don't understand the final equation, would you please help? You mentioned to replace the secord term with (x-1), but it appeared as -(x+1) . Also, removing the paranthesis leaves you with 2x from the "+(x+5)+(x-5)", how did that work?
Thanks!



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