# Math

posted by on .

Solve the inequality:

2x-1 / 3x-2 < 1
[the < is actually less than or equal to]

I got one of the answers, x is less than or equal to 1, but there's another. How do you find that answer?

Thanks

• Math - ,

I mean x is greater than or equal to, not less than. sorry.

• Math - ,

I got:

x < 2/3 OR x >= 1

The way I got it was not straightfoward, but there seems to have been method in my madness since my solution seems to work. I checked my answer for some key and other values (x = 2/3, x = 1, x = 3/4, x = 5/6, x = 10, x = -10) and all of them turned out right.

Before we go any farther we need to ruleout the special case where the denominator = 0. That is, we solve denominator = 0 to find the illegal values for x. 3x - 2 = 0 leads to x = 2/3. So when x = 2/3, the denominator is 0, which is undefined. Thus, 2/3 cannot be part of any solution.

Okay, the off-beat part was having to handle the 2 possible cases when multiplying through by 3x - 2 to clear the inequality of fractions. If, for an inequality, you multiply by a negative number you have to reverse the direction of the inequality sign: if you multiply by a positive number, you don't.

Well, is 3x - 2 positive or negative? We don't know. So we have to try both.

I split the problem into 2 separate inequalities: one for 3x - 2 > 0 and one for 3x - 2 < 0. In simplified form they are x > 2/3 and x < 2/3, respectively.

Solving for the first I got x >= 1. Solving for the second I got x <= 1.

Plotting those answers on a number line shows that ALL (real) numbers are a solution (except for 2/3, which we ruled out at the beginning).

That seems fine, but wait. Consider the first branch, where x > 2/3. The answer obtained is x >= 1. So IF x > 2/3, then x >= 1????? What if x = 4/5? Then it is greater than 2/3 but less than 1. A contradiction? Plugging 3/4 in for x I got an invalid solution. So x = 3/4 is not a solution. In fact, any value for x that is greater than 2/3 but less than 1 is not a solution.

Next, I considered the second branch, where x < 2/3. The answer obtained is x <= 1. Looking at it the problem region matched that of the first branch.

Therefore, all numbers are a solution except for 2/3 (which would make the denominator 0) and any numbers greater than 2/3 and less than 1.

Consequently:

x < 2/3 OR x >= 1