Absolute Value with Inequalities

[LisaPersad]

If the rule is the following:

I ax+b I > k is = to ax + b > k or ax + b < -k

Then why in this problem below do we solve it differently? Than what the rule suggests?

The problem:
I 7x + 3 I < 25

-25 < (7x + 3) < 25
(-25 - 3) < (7x) < (25 - 3)
(-28) < (7x) < (22)
(-28/7) < (7x/7) < (22/7)
(-4 < x < 22/7)

The solution interval form is (-4, 22/7)

 

[Deleted member 4993]

Absolute Value with Inequalities
by LisaPersad on Fri Feb 27, 2009 4:17 pm

If the rule is the following:

I ax+b I > k is = to ax + b > k and ax + b < -k

Then why in this problem below do we solve it differently? Than what the rule suggests?

Did you try to follow the rule above and come up with different answer? If you did post your work and we will check it out.

 

[stapel]

"Greater than" absolute-value inequalities work differently from "less than" absolute-value inequalities!

For an explanation, try here. :wink:

 

[LisaPersad]

OMG!!! I got the same answer!!! As before!!!! That's great!!! :lol:

The problem: I 7x + 3 I < 25 can be worked the other way too!!!
As follows:
7x + 3 < 25 OR 7x + 3 > -25
7x < 25-3 7x > -25-3
7x < 22 7x/7 > -28/7
7x/7 < 22/7 x> - 4
x< 22/7

Thank you soooo much for your help!!!!!