Fractions marred in inequality

9/6/2010 11:49:53 PM #

It has been a while since I dealt with these things, but let's start here:
There's a boundary between this being true and being false. That boundary is at:

(10x ÷ (2x + 3)) = 5

If we solve for that, then go back and figure out which side satisfies the inequality, that will save us having to deal with the inequality for the whole time.

As you did above, we go along …

10x = 5 (2x + 3)
10x = 10x + 15
0 = 15
false.
So I would have to guess that this thing is never true, but as I said, it's been a while, so I'll keep thinking.

I was also going to worry about when (2x + 3) is zero, because dividing by that will cause trouble, but that only happens at x=-1.5 and should be fine everywhere else.

The other thing to worry about is when (2x + 3) is negative, because multiplying both sides by that will require reversing the inequality. Aha! Give that a try: whenever you multiply by a term that's (even partially) unknown, you have to split your reasoning into two paths, for "if the new factor is positive" and "if it's negative".

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