Ed 5, Algebra, Extra Inequalities Problem Set # 3

[harsh.vinayak]

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

I understand I am wrong here but the way I naturally would solve the given equation is:
4/x - 1/3 < 0
(12 - x)/(3x) < 0

Two cases:
a) 12 - x < 0 & 3x > 0 :: x > 12 & x > 0 therefore x > 12
b) 12 - x > 0 & 3x < 0 :: x < 12 & x < 0 therefore x < 12

but the solution states x < 0 or x > 12 . There is also an errata in the solution which I have reported. Can someone kindly explain?

[jnelson0612]

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

I understand I am wrong here but the way I naturally would solve the given equation is:
4/x - 1/3 < 0
(12 - x)/(3x) < 0

Two cases:
a) 12 - x < 0 & 3x > 0 :: x > 12 & x > 0 therefore x > 12
b) 12 - x > 0 & 3x < 0 :: x < 12 & x < 0 therefore x < 12

but the solution states x < 0 or x > 12 . There is also an errata in the solution which I have reported. Can someone kindly explain?

Let's go back to your two cases:

Two cases:
a) 12 - x < 0 & 3x > 0 :: x > 12 & x > 0 therefore x > 12
Correct! If x is both greater than 0 and greater than 12, then x must be greater than 12.

Here's your mistake:
b) 12 - x > 0 & 3x < 0 :: x < 12 & x < 0 therefore x < 12
Not quite. If x is less than 12 AND less than 0 then x must be less than 0, not 12. If you have any doubt, graph both of these on the number line and look at where the overlap is. We only see overlap once x is less than 0.

[harsh.vinayak]

Thanks Jamie :) . I also found the graphical approach to solve this question very convenient ..

[tim]

:)