Answer
$(-\infty, -1] \cup \left(\frac{5}{2}, 3\right]$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a Rational function.
$\displaystyle x\leq\frac{6-x}{2x-5},$
$\displaystyle x-\frac{6+x}{2x-5}\leq0,$
$\displaystyle \frac{x(2x-5)-6+x}{2x-5}\leq0,$
$\displaystyle \frac{2x^2-4x-6}{2x-5}\leq0,$
$f(x)=\displaystyle \frac{(2x-6)(x+1)}{2x-5}\leq0,$
2. The cut points are:
$\displaystyle\frac{(2x-6)(x+1)}{2x-5} \lt 0$
$x=-1$ or $x=\frac{5}{2}$ or $x=3$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \leq 0 ? \\
& & \frac{(2a-6)(a+1)}{2a-5} & \\
(-\infty,-1) & -5 & \frac{(-)(-)}{(-)}=(-) & T\\
(-1,\frac{5}{2}) & 0 & \frac{(-)(+)}{(-)}=(+) & F\\
(\frac{5}{2}, 3) & \frac{8}{3} & \frac{(-)(+)}{(+)}=(-) & T\\
(3,\infty) & 5 & \frac{(+)(+)}{(+)}=(+) & F\\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-\infty, -1] \cup \left(\frac{5}{2}, 3\right]$
