College Algebra, Chapter 1, 1.6, Section 1.6, Problem 60

Solve the nonlinear inequality $\displaystyle -2 < \frac{x+1}{x-3} $. Express the solution using interval notation and graph the solution set.

$
\begin{equation}
\begin{aligned}
-2 & < \frac{x+1}{x-3}\\
\\
0 & < \frac{x+1}{x-3} +2 && \text{Add } 2\\
\\
0 & < \frac{x+1+2x-6}{x-3} && \text{Common denominator}\\
\\
0 & < \frac{3x-5}{x-3}
\end{aligned}
\end{equation}
$


The factors on the left hand side are $3x-5$ and $x-3$. These factors are zero when $x$ is $\displaystyle \frac{5}{3}$ and 3 respectively. These numbers divide the real line into intervals
$\left( -\infty, \frac{5}{3} \right), \left( \frac{5}{3},3 \right), (3,\infty)$




From the diagram, the solution of the inequality $\displaystyle 0 < \frac{3x-5}{x-3}$ are
$\left( -\infty, \frac{5}{3} \right) \bigcup (3, \infty)$

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