Find all solutions to the inequality \[\frac{x}{x-1} + \frac{x+2}{2x} \ge 3.\](Give your answer in interval notation.)
Subtracting $3$ from both sides gives \[\frac{x}{x-1} + \frac{x+2}{2x} -3 \ge 0.\]Combining all the terms under a common denominator, we get \[\frac{x(2x) + (x+2)(x-1) - 3(x-1)(2x)}{(x-1)(2x)} \ge 0,\]or \[\frac{-3x^2+7x-2}{2x(x-1)} \ge 0.\]Factoring the numerator, we get \[\frac{-(3x-1)(x-2)}{2x(x-1)} \ge 0.\]Making a sign table for the inequality $f(x) = \frac{(3x-1)(x-2)}{x(x-1)} \le 0,$ we get:  \begin{tabular}{c|cccc|c} &$3x-1$ &$x-2$ &$x$ &$x-1$ &$f(x)$ \\ \hline$x<0$ &$-$&$-$&$-$&$-$&$+$\\ [.1cm]$0<x<\frac{1}{3}$ &$-$&$-$&$+$&$-$&$-$\\ [.1cm]$\frac{1}{3}<x<1$ &$+$&$-$&$+$&$-$&$+$\\ [.1cm]$1<x<2$ &$+$&$-$&$+$&$+$&$-$\\ [.1cm]$x>2$ &$+$&$+$&$+$&$+$&$+$\\ [.1cm]\end{tabular}Therefore, we have $f(x) < 0$ when $0 < x < \tfrac13$ or $1 < x <2.$ We also have $f(x) = 0$ when $x = \tfrac13$ or $x = 2,$ so the whole solution set to the inequality is \[x \in \boxed{(0, \tfrac13] \cup (1, 2]}.\]
(0,\tfrac{1}{3}]\cup(1,2]
Level 4
Intermediate Algebra
Problem: Find all solutions to the inequality \[\frac{x}{x-1} + \frac{x+2}{2x} \ge 3.\](Give your answer in interval notation.)
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the inequality, first, combine the fractions on the left side. The common denominator is \(2x(x-1)\). Rewrite the left side as:

\[
\frac{x \cdot 2x + (x+2)(x-1)}{2x(x-1)} \ge 3.
\]

Expanding the numerator gives:

\[
\frac{2x^2 + (x^2 + 2x - x - 2)}{2x(x-1)} = \frac{2x^2 + x^2 + x - 2}{2x(x-1)} = \frac{3x^2 + x - 2}{2x(x-1)} \ge 3.
\]

Next, subtract 3 from both sides to set the inequality to zero:

\[
\frac{3x^2 + x - 2}{2x(x-1)} - 3 \ge 0.
\]

This simplifies to:

\[
\frac{3x^2 + x - 2 - 6x(x-1)}{2x(x-1)} \ge 0.
\]

Expanding \(6x(x-1)\) results in:

\[
6x^2 - 6x.
\]

Thus, the inequality becomes:

\[
\frac{3x^2 + x - 2 - 6x^2 + 6x}{2x(x-1)} \ge 0,
\]

which simplifies to:

\[
\frac{-3x^2 + 7x - 2}{2x(x-1)} \ge 0.
\]

Next, factor the quadratic in the numerator. The roots can be found using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{49 - 24}}{-6} = \frac{-7 \pm 5}{-6}.
\]

Calculating the roots gives:

\[
x = \frac{-2}{-6} = \frac{1}{3} \quad \text{and} \quad x = \frac{-12}{-6} = 2.
\]

Thus, the quadratic factors as:

\[
-3(x - \frac{1}{3})(x - 2).
\]

The inequality now reads:

\[
\frac{-3(x - \frac{1}{3})(x - 2)}{2x(x-1)} \ge 0.
\]

Identify critical points: \(x = 0\), \(x = 1\), \(x = \frac{1}{3}\), and \(x = 2\). Analyze the sign of the expression in each interval determined by these points.

1. For \(x < 0\), the expression is negative.
2. For \(0 < x < \frac{1}{3}\), the expression is negative.
3. For \(\frac{1}{3} < x < 1\), the expression is positive.
4. For \(1 < x < 2\), the expression is negative.
5. For \(x > 2\), the expression is positive.

Now, check the endpoints. At \(x = \frac{1}{3}\) and \(x = 2\), the expression equals zero, so these points are included in the solution. At \(x = 0\) and \(x = 1\), the expression is undefined.

Thus, the solution in interval notation is:

\[
(0, \frac{1}{3}] \cup (1, 2].
\]<|END_OF_LATENT|> The final answer is $(0,\tfrac{1}{3}]\cup(1,2]$.