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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.
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.
19 Aug 2008, 09:46
Kudos
Bookmarks
amitdgr
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.
