College Algebra, Chapter 1, 1.6, Section 1.6, Problem 64

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

$
\begin{equation}
\begin{aligned}
\frac{x}{x+1} & > 3x\\
\\
\frac{x}{x+1} - 3x & > 0 && \text{Subtract } 3x\\
\\
\frac{x-3x^2-3x }{x +1 } & > 0 && \text{Common denominator}\\
\\
\frac{-3x^2-2x}{x+1}& > 0 && \text{Simplify the numerator}\\
\\
\frac{-x(3x+2)}{x+1} & > 0 && \text{Factor out} -x\\
\\
\frac{x(3x+2)}{x+1} & < 0 && \text{Divide by } -1
\end{aligned}
\end{equation}
$


The factors on the left hand side are $x$, $3x+2$ and $x+1$. These factors are zero when $x$ is 0, $\displaystyle \frac{-2}{3}$ and -1 respectively. These numbers divide the real line into intervals
$(-\infty, -1) \left( -1, -\frac{2}{3}\right), \left( -\frac{2}{3}, 0\right), (0,\infty)$




From the diagram, the solution of the inequality $\displaystyle \frac{x(3x+2)}{x+1} < 0$ are
$\displaystyle (-\infty,-1) \bigcup \left( -\frac{2}{3}, 0 \right)$