Print Page - absolute value inequalites

Title: absolute value inequalites
Post by: LivedKrad on October 21, 2004, 06:05 PM

I'm having trouble following procedure for solving absolute value inequalities.

e.g. | (( x + 5 ) / 3 ) | - 3 > 6

I proceded to set the inequality up as two inequalities..

(( x + 5) / 3 ) - 3 < -6 or (( x + 5) / 3 ) - 3 > 6

I then solved accordingly...

(( x + 5) / 3 ) < -3 or (( x + 5) / 3 ) > 9

(x + 5) < -9 or (x + 5) > 27

x < -14 or x > 22

Yet, this is wrong! I can't seem to figure out what I did wrong. Help please :(

Title: Re: absolute value inequalites
Post by: LivedKrad on October 21, 2004, 06:32 PM

UPDATE: Solved.

Title: Re: absolute value inequalites
Post by: j0k3r on October 21, 2004, 07:11 PM

Generally when you have a problem and solve it, you post your solution so that others may learn from it.

Title: Re: absolute value inequalites
Post by: Yoni on October 22, 2004, 04:59 AM

You can't treat the -3 with such disrespect.
First it has to blend in with the 6 in the RHS.

| (( x + 5 ) / 3 ) | > 9

Now proceed normally...

Valhalla Legends Archive

Member Forums => Yoni's Math Forum => Topic started by: LivedKrad on October 21, 2004, 06:05 PM