inequality question ..please help

pcmdotcom wrote:If 4/x < 1/3, what is the possible range of values for x?

Answer [spoiler]x<0 and x>12[/spoiler]

x can be either positive or negative.

Case 1: x is positive
4/x < 1/3
x > 12

Case 2: x is negative => x = -m _______m is a positive number
4/(-m) < 1/3
-4/m < 1/3
This is true for all positive numbers m.
So it is true for all negative numbers x.
x < 0

So [spoiler]x < 0[/spoiler] and [spoiler]x > 12[/spoiler]

pcmdotcom wrote:Also

If 4/x < -1/3, what is the possible range of values for x?
Answer [spoiler]-12<x<0[/spoiler]

Please explain how these answers have been reached. Thank you in advance

Case 1: x > 0
4/x < -1/3
This will never be true.

Case 2: x < 0
Let x = -m _______m is a positive number
4/(-m) < -1/3
-4/m < -1/3
4/m > 1/3
m < 12
(-m) > -12
x > -12

So [spoiler]-12 < x < 0[/spoiler]

Cheers

pcmdotcom wrote:If 4/x < 1/3, what is the possible range of values for x?

Answer x<0 and x>12

Also

If 4/x < -1/3, what is the possible range of values for x?
Answer -12<x<0

Please explain how these answers have been reached. Thank you in advance

Hi Pcmdotcom,
For the first One,
4/x < 1/3

Divide by 4 on both sides,
1/x < 1/12

Case 1: LHS can be less than RHS when x takes the negative values. So X<0
(When X takes negative LHS will be negative irrespective of value x and RHS will be positive.)
Case 2: When X takes positive value, RHS will be greater than LHS only when X>12
Reason,
Since both the sign is positive and numerator is same, Then Denominator with Higher value will be the smallest one.
Example: 1/100 < 1/10 (0.01<0.1)
If sign is negative it will be the opposite case:
Example: -(1/100) > -(1/10) (-0.01 > -0.1)

Hope it helps you.

Regards,
Uva.

For Question 1:

4/x < 1/3

If "x" is positive .. cross multiply...

x > 12.... So overlap is x>12

if "x" is negative.. multiply both sides by "-1" .. so sign changes.

x < 12.... So overlap is x<0

pcmdotcom wrote:If 4/x < 1/3, what is the possible range of values for x?

The CRITICAL POINTS are where 4/x = 1/3 and where 4/x is undefined.
4/x = 1/3 when x=12.
4/x is undefined when x=0.
To determine the ranges where 4/x < 1/3, test one value to the left and right of each critical point.

x<0:
Plugging x=-1 into 4/x < 1/3, we get:
4/-1 < 1/3
-4 < 1/3.
This works.
Thus, x<0 is a valid range.

0<x<12:
Plugging x=1 into 4/x < 1/3, we get:
4/1 < 1/3
4 < 1/3.
Doesn't work.
Thus, 0<x<12 is not a valid range.

x>12:
Plugging x=16 into 4/x < 1/3 , we get:
4/16 < 1/3
1/4 < 1/3.
This works.
Thus, x>12 is a valid range.

Result:
4/x < 1/3 when x<0 or x>12.

4/x < 1/3
Multiply by 3x to get:
12 < x
So x > 12

Also, dividing 4 by a negative value will create a negative result (implicitly < 13)
Dividing by 0 is not possible.

Answer x<0 and x>12

If 4/x < 1/3, what is the possible range of values for x?

Answer x<0 and x>12

The x<0 is not easy to spot.