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Points to remember : 1. if x > y then (1/x) < (1/y) (

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Points to remember : 1. if x > y then (1/x) < (1/y) ( [#permalink] New post 18 Aug 2008, 01:44
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Points to remember :
1. if x > y then (1/x) < (1/y) ( EDIT:Holds good only when BOTH x and Y share SAME sign - Thanks Ian)
2. if x > y then -x < -y (that is the inequality gets reversed when both sides are multiplied by negative sign)
3. NEVER EVER , cross-multiply a variable (or) expression in an inequality blindly. You CAN cross multiply if and only if you are sure that the variable (or) expression being cross-multiplied is positive.
4. You can blindly cross multiply constant terms or numbers.

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

For any real numbers, a; b,and c:
a < b is equivalent to a + c < b + c;
a > b is equivalent to a + c > b + c;
a = b is equivalent to a + c = b + c;
a >= b is equivalent to a + c >= b + c.
a <= b is equivalent to a + c <= b + c.

In other words, when we add or subtract the same number on both sides of an inequality, the direction of the inequality symbol is not changed.

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

For any real numbers, a; b, and any positive number c:
a < b is equivalent to ac < bc;
a > b is equivalent to ac > bc.
a < b is equivalent to a/c < b/c;
a > b is equivalent to a/c > b/c.

For any real numbers, a; b, and any negative number c:
a < b is equivalent to ac > bc;
a > b is equivalent to ac < bc.
a < b is equivalent to a/c > b/c;
a > b is equivalent to a/c < b/c.

Similar statements hold for >= and <=

In other words, when we multiply or divide by a positive number on both sides of an inequality, the direction of the inequality symbol stays the same.

When we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol is reversed.

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

|x| = x if x > 0
|x| = -x if x < 0
if |x| > y , then either x > y or -x > y
if |x| < y , then either x < y or -x < y

let "r" be a positive real number and "a" be a fixed real number, then
|x-a| < r implies a-r < x < a+r in other words x lies somewhere in between a-r and a+r
|x-a| > r implies x < a-r or x > a+r in other words, x lies outside a+r and a-r

----------------------------------------------------------------------------------------------------------------------------------
if k is a positive integer:

1 )if x^2 > k^2 and x> 0 then this implies x > k
2) k^x > 1 when x>0
3) 0< k^x < 1 when x < 0

------------------------------------------------------------------------------------------------------------------------------------
For word problems :

x is at least 30 implies x>=30 ( that is x is minimum 30)
x is at most 30 implies x<=30 ( that is x is maximum 30)
x cannot exceed 45 implies x <=45 ( that is x is maximum 45)
x must exceed 34 implies x > 34
x is between 7 and 12 implies 7 < x < 12
------------------------------------------------------------------------------------------------------------------------------

Please add if i missed something.
_________________

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Last edited by amitdgr on 19 Sep 2008, 05:02, edited 3 times in total.
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Re: Inequalities Review [#permalink] New post 19 Aug 2008, 02:10
amitdgr, an excellent compilation.
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Re: Inequalities Review [#permalink] New post 19 Aug 2008, 08:46
amitdgr wrote:
Points to remember :

1. if x > y then (1/x) < (1/y)
2. if x > y then -x < -y (that is the inequality gets reversed when both sides are multiplied by negative sign)
3. NEVER EVER , cross-multiply a variable (or) expression in an inequality blindly. You CAN cross multiply if and only if you are sure that the variable (or) expression being cross-multiplied is positive.
4. You can blindly cross multiply constant terms or numbers.
--------------------------------------------------------------------------------------------------------------------------

For any real numbers, a; b,and c:
a < b is equivalent to a + c < b + c;
a > b is equivalent to a + c > b + c;
a = b is equivalent to a + c = b + c;
a >= b is equivalent to a + c >= b + c.
a <= b is equivalent to a + c <= b + c.

In other words, when we add or subtract the same number on both sides of an inequality, the direction of the inequality symbol is not changed.

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

For any real numbers, a; b, and any positive number c:
a < b is equivalent to ac < bc;
a > b is equivalent to ac > bc.
a < b is equivalent to a/c < b/c;
a > b is equivalent to a/c > b/c.

For any real numbers, a; b, and any negative number c:
a < b is equivalent to ac > bc;
a > b is equivalent to ac < bc.
a < b is equivalent to a/c > b/c;
a > b is equivalent to a/c < b/c.

Similar statements hold for >= and <=

In other words, when we multiply or divide by a positive number on both sides of an inequality, the direction of the inequality symbol stays the same.

When we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol is reversed.

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

|x| = x if x > 0
|x| = -x if x < 0
if |x| > y , then either x > y or -x > y
if |x| < y , then either x < y or -x < y

let "r" be a positive real number and "a" be a fixed real number, then
|x-a| < r implies a-r < x < a+r in other words x lies somewhere in between a-r and a+r
|x-a| > r implies x < a-r or x > a+r in other words, x lies outside a+r and a-r

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

1 )if x^2 > k^2 and x> 0 then this implies k >0

2) If k > 1 , then

a) k^x > 1 when x>0
b) 0< k^x < 1 when x < 0

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

For word problems :

x is at least 30 implies x>=30 ( that is x is minimum 30)
x is at most 30 implies x<=30 ( that is x is maximum 30)
x cannot exceed 45 implies x <=45 ( that is x is maximum 45)
x must exceed 34 implies x > 34
x is between 7 and 12 implies 7 < x < 12
------------------------------------------------------------------------------------------------------------------------------


Please add if i missed something.
_________________
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Thanks for posting the summary.

How is this possible?
1 ) if x^2 > k^2 and x> 0 then this implies k >0
We can have x=2 and k = -1 and still satisfy the condition x^2 > k^2; however, in this case k !>0.
I think we can say that based on the given conditions, x > k.
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Re: Inequalities Review [#permalink] New post 19 Aug 2008, 09:12
leonidas wrote:
How is this possible?
1 ) if x^2 > k^2 and x> 0 then this implies k >0
We can have x=2 and k = -1 and still satisfy the condition x^2 > k^2; however, in this case k !>0.
I think we can say that based on the given conditions, x > k.


I don't think this is always true, especially since this doesn't limit the rule to positive integers.

x = -2
k = 1/2

x^2 = 4
k^2 = 1/4

x !> k
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Re: Inequalities Review [#permalink] New post 18 Sep 2008, 08:41
jallenmorris wrote:
leonidas wrote:
How is this possible?
1 ) if x^2 > k^2 and x> 0 then this implies k >0
We can have x=2 and k = -1 and still satisfy the condition x^2 > k^2; however, in this case k !>0.
I think we can say that based on the given conditions, x > k.


I don't think this is always true, especially since this doesn't limit the rule to positive integers.

x = -2
k = 1/2

x^2 = 4
k^2 = 1/4

x !> k


Nice catch....... you got me there!!!!! I agree, I assumed them to be integers (though not given), got side tracked..
The above should hold good, if they were integers.
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Re: Inequalities Review [#permalink] New post 18 Sep 2008, 10:47
amitdgr wrote:
Points to remember :
1. if x > y then (1/x) < (1/y)
Please add if i missed something.


I didn't go through this post thoroughly, but 1 above is not quite right. Notice that if x = 3 and y = -2, for example, then x > y and 1/x is also greater than 1/y.

1 above is only true if x and y share the same sign, as one can quite easily demonstrate (start from x > y, then divide both sides by x and by y).
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Re: Inequalities Review [#permalink] New post 22 Jul 2009, 21:11
Great compilation. Thanks!!!
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Re: Inequalities Review [#permalink] New post 22 Jul 2009, 22:55
Nice compilation, Thanks.
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Re: Inequalities Review [#permalink] New post 10 Nov 2009, 14:29
amitdgr wrote:
----------------------------------------------------------------------------------------------------------------------------------
if k is a positive integer:

1 )if x^2 > k^2 and x> 0 then this implies x > k
2) k^x > 1 when x>0
3) 0< k^x < 1 when x < 0

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


Hi. Appreciate the effort! Good Job!

Just a minor correction though. Point 2 and 3 (in red above) are not valid for k = 1.

Cheers.
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Re: Inequalities Review [#permalink] New post 11 Nov 2009, 19:59
amitdgr wrote:
Points to remember :
1. if x > y then (1/x) < (1/y) ( EDIT:Holds good only when BOTH x and Y share SAME sign - Thanks Ian)
2. if x > y then -x < -y (that is the inequality gets reversed when both sides are multiplied by negative sign)
3. NEVER EVER , cross-multiply a variable (or) expression in an inequality blindly. You CAN cross multiply if and only if you are sure that the variable (or) expression being cross-multiplied is positive.
4. You can blindly cross multiply constant terms or numbers.

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

For any real numbers, a; b,and c:
a < b is equivalent to a + c < b + c;
a > b is equivalent to a + c > b + c;
a = b is equivalent to a + c = b + c;
a >= b is equivalent to a + c >= b + c.
a <= b is equivalent to a + c <= b + c.

In other words, when we add or subtract the same number on both sides of an inequality, the direction of the inequality symbol is not changed.

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

For any real numbers, a; b, and any positive number c:
a < b is equivalent to ac < bc;
a > b is equivalent to ac > bc.
a < b is equivalent to a/c < b/c;
a > b is equivalent to a/c > b/c.

For any real numbers, a; b, and any negative number c:
a < b is equivalent to ac > bc;
a > b is equivalent to ac < bc.
a < b is equivalent to a/c > b/c;
a > b is equivalent to a/c < b/c.

Similar statements hold for >= and <=

In other words, when we multiply or divide by a positive number on both sides of an inequality, the direction of the inequality symbol stays the same.

When we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol is reversed.

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

|x| = x if x > 0
|x| = -x if x < 0
if |x| > y , then either x > y or -x > y
if |x| < y , then either x < y or -x < y

let "r" be a positive real number and "a" be a fixed real number, then
|x-a| < r implies a-r < x < a+r in other words x lies somewhere in between a-r and a+r
|x-a| > r implies x < a-r or x > a+r in other words, x lies outside a+r and a-r

----------------------------------------------------------------------------------------------------------------------------------
if k is a positive integer:

1 )if x^2 > k^2 and x> 0 then this implies x > k
2) k^x > 1 when x>0
3) 0< k^x < 1 when x < 0

------------------------------------------------------------------------------------------------------------------------------------
For word problems :

x is at least 30 implies x>=30 ( that is x is minimum 30)
x is at most 30 implies x<=30 ( that is x is maximum 30)
x cannot exceed 45 implies x <=45 ( that is x is maximum 45)
x must exceed 34 implies x > 34
x is between 7 and 12 implies 7 < x < 12
------------------------------------------------------------------------------------------------------------------------------

Please add if i missed something.


Thanks @amitdgr +1 from me. Great collection
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Re: Inequalities Review [#permalink] New post 20 Jan 2010, 01:11
amitdgr wrote:
Points to remember :
1. if x > y then (1/x) < (1/y) ( EDIT:Holds good only when BOTH x and Y share SAME sign - Thanks Ian)
2. if x > y then -x < -y (that is the inequality gets reversed when both sides are multiplied by negative sign)
3. NEVER EVER , cross-multiply a variable (or) expression in an inequality blindly. You CAN cross multiply if and only if you are sure that the variable (or) expression being cross-multiplied is positive.
4. You can blindly cross multiply constant terms or numbers.

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

For any real numbers, a; b,and c:
a < b is equivalent to a + c < b + c;
a > b is equivalent to a + c > b + c;
a = b is equivalent to a + c = b + c;
a >= b is equivalent to a + c >= b + c.
a <= b is equivalent to a + c <= b + c.

In other words, when we add or subtract the same number on both sides of an inequality, the direction of the inequality symbol is not changed.

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

For any real numbers, a; b, and any positive number c:
a < b is equivalent to ac < bc;
a > b is equivalent to ac > bc.
Quote:
a < b is equivalent to a/c < b/c;

a > b is equivalent to a/c > b/c.

For any real numbers, a; b, and any negative number c:
a < b is equivalent to ac > bc;
a > b is equivalent to ac < bc.
a < b is equivalent to a/c > b/c;
a > b is equivalent to a/c < b/c.

Similar statements hold for >= and <=

In other words, when we multiply or divide by a positive number on both sides of an inequality, the direction of the inequality symbol stays the same.

When we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol is reversed.

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

|x| = x if x > 0
|x| = -x if x < 0
if |x| > y , then either x > y or -x > y
if |x| < y , then either x < y or -x < y

let "r" be a positive real number and "a" be a fixed real number, then
|x-a| < r implies a-r < x < a+r in other words x lies somewhere in between a-r and a+r
|x-a| > r implies x < a-r or x > a+r in other words, x lies outside a+r and a-r

----------------------------------------------------------------------------------------------------------------------------------
if k is a positive integer:

1 )if x^2 > k^2 and x> 0 then this implies x > k
2) k^x > 1 when x>0
3) 0< k^x < 1 when x < 0

------------------------------------------------------------------------------------------------------------------------------------
For word problems :

x is at least 30 implies x>=30 ( that is x is minimum 30)
x is at most 30 implies x<=30 ( that is x is maximum 30)
x cannot exceed 45 implies x <=45 ( that is x is maximum 45)
x must exceed 34 implies x > 34
x is between 7 and 12 implies 7 < x < 12
------------------------------------------------------------------------------------------------------------------------------

Please add if i missed something.


This wont be true if c is between 0 and 1. a < b is equivalent to a/c < b/c
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Re: Inequalities Review [#permalink] New post 30 Jan 2011, 23:47
@ above post ,

a/c < b/c , if a < b and 0 < c < 1 , of course even in the given constrains its true.

thanks for the help
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Re: Inequalities Review [#permalink] New post 06 Jun 2011, 19:51
Hi,

can someone please explain this :

let "r" be a positive real number and "a" be a fixed real number, then
|x-a| < r implies a-r < x < a+r in other words x lies somewhere in between a-r and a+r
|x-a| > r implies x < a-r or x > a+r in other words, x lies outside a+r and a-r


How do we open the modulus here?

I came across this example :

If |5x + 2| – 3 < 3x, what is the value of x?
|5x+2|< 3x+3 |5x+2| < -3x-3

2x < 1 5x+2 < -3x - 3
x< 1/2 8x < -5 ---> x < -5/8


so,if we open |x-a| < r as the above:

|x-a| < r |x-a| < -r
x < r+a x < a-r


What am i missing :?



regards,
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Re: Inequalities Review [#permalink] New post 10 Jun 2011, 19:22
Excellent compilation...good refresher
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Re: Inequalities Review   [#permalink] 10 Jun 2011, 19:22
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