College Algebra, Chapter 1, 1.6, Section 1.6, Problem 66

Solve the nonlinear inequality $\displaystyle \frac{3}{x-1} - \frac{4}{x} \geq 1 $. Express the solution using interval notation and graph the solution set.

$
\begin{equation}
\begin{aligned}
\frac{3}{x-1} - \frac{4}{x} \geq 1\\
\\
\frac{3}{x-1} - \frac{4}{x} - 1 & \geq 0 && \text{Subtract } 1\\
\\
\frac{3x-4(x-1)-x(x-1)}{x(x-1)} & \geq 0 && \text{Multiply LCD } x(x-1)\\
\\
\frac{3x-4x+4-x^2+x}{x(x-1)} & \geq 0 && \text{Simplify the numerator}\\
\\
\frac{-x^2+4}{x(x-1)} & \geq 0&& \text{Factor out negative } 1\\
\\
\frac{-(x^2-4)}{x(x-1)} & \geq 0 && \text{Divide by } -1 \\
\\
\frac{x^2 - 4}{x(x-1)} & \leq 0 && \text{Difference of squares}\\
\\
\frac{(x+2)(x-2)}{x(x-1)} & \leq 0

\end{aligned}
\end{equation}
$


The factors on the left hand side are $x$, $x+2$, $x-2$ and $x-1$. These factors are zero when $x$ is 0,-2,2 and 1 respectively. These numbers divide the real line into intervals
$(-\infty, -2], (-2,0), (0,1),(1,2),[2,\infty)$




From the diagram, the solution of the inequality $\displaystyle \frac{(x+2)(x-2)}{x(x-1)} \leq 0$ are
$[-2,0) \bigcup (1,2]$