S.O.S. Mathematics CyberBoard

I am curious about your sites solution to inequalities in the form

|ax + b| < cx + d

You apply the definition of absolute value to state

(ax + b > 0 and ax + b < cx + d) or (ax + b < 0 and -(ax + b) < cx + d)

After solving the conjunctions, the union of the results produces the answer.

I am looking for a case where this would omit any extraneous solutions produced by assuming

-(cx + d) < ax + b < cx + d (similar to your proposed solution to |ax + b| < c)

but have not been able to find one. Would this assumption be valid for linear expressions?

It's valid for all such. The two cases or the one case listed at the same time, you get the same answers.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Why would we need to go through the longer discussion if this is the case?

It's a matter of style, some people like cases, some people like doing them all at once, it's a matter of preference. Think of it as texting someone versus calling them. Both can convey information, but you make a choice as to which you want to use at the time, as doing both would certainly be wasteful.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Another reason is that splitting into cases works for many other inequalities, such as (or with ) for some functions p,q,r, where there isn't a way to do them all at once without introducing erroneous solutions that you will need to investigate later.

_________________