College Algebra, Chapter 1, 1.6, Section 1.6, Problem 62

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

$
\begin{equation}
\begin{aligned}
\frac{3+x}{3-x} & \geq 1\\
\\
\frac{3+x}{3-x} - 1 & \geq 0 && \text{Subtract } 1\\
\\
\frac{3+x}{3-x} - \left( \frac{3-x}{3-x} \right) & \geq 0 && \text{Common Denominator}\\
\\
\frac{3+x-3+x}{3-x} & \geq 0 && \text{Simplify}\\
\\
\frac{2x}{3-x} & \geq 0
\end{aligned}
\end{equation}
$


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




From the diagram, the solution of the inequality $\displaystyle \frac{2x}{3-x} \geq 0$ are
$[0,3)$