The expression (2 - 3x)/(x(x - 1)) is positive. To which set of values does x belong?

Step-by-step solution

1. 1

   List the sign-change points. A quotient changes sign only where a factor is zero. Numerator: $2-3x = 0$ at $x = \tfrac23$. Denominator: $x = 0$ and $x = 1$. So three points, in the order $0 < \tfrac23 < 1$.

2. 2

   Note that the two kinds of point behave differently. At $x = \tfrac23$ the fraction equals $0$, which is not positive, so $\tfrac23$ is excluded from a strict inequality. At $x = 0$ and $x = 1$ the fraction is undefined, so those are excluded as well. Every endpoint here is open.

3. 3

   Test each of the four intervals.

   interval	test $x$	numerator	denominator	quotient
   $(-\infty,0)$	$-1$	$+5$	$(-1)(-2) = +2$	positive
   $(0,\tfrac23)$	$0.5$	$+0.5$	$(0.5)(-0.5) = -0.25$	negative
   $(\tfrac23,1)$	$0.8$	$-0.4$	$(0.8)(-0.2) = -0.16$	positive
   $(1,\infty)$	$2$	$-4$	$(2)(1) = +2$	negative

4. 4

   Read off the positive intervals. The quotient is positive on the first and third:

   $x \in (-\infty,\,0) \cup \left(\tfrac23,\,1\right)$

5. 5

   Understand why the signs alternate the way they do. Crossing $x = 0$ or $x = 1$ flips one linear factor in the denominator, and crossing $x = \tfrac23$ flips the numerator — so the sign alternates at every one of the three points, which is exactly the pattern in the table.

6. 6

   Confirm symbolically. A solver applied to $\dfrac{2-3x}{x(x-1)} > 0$ returns $(-\infty,0) \cup \left(\tfrac23,1\right)$ ✓, matching the hand sign chart. The tempting wrong answer $(-\infty,0) \cup (\tfrac23,\infty)$ forgets that the fraction turns negative again past $x = 1$.