3. All the values of $x$ satisfying the inequality

$$\frac{1}{x+5}\le \frac{1}{2x-3}$$

are

$$\frac{\mathrm{}}{}$$

The correct answer is:
$x<-5$ or $x>\frac{3}{2}$

Step-by-Step Explanation:

To find the values of $x$ that satisfy the inequality:
$$\frac{1}{x+5}\le \frac{1}{2x-3}$$
we must first identify the values of $x$ where the expressions are undefined. These are the points where the denominators equal zero.

1. Set the first denominator to zero:
$x+5=0\Rightarrow x=-5$
2. Set the second denominator to zero:
$2x-3=0\Rightarrow x=\frac{3}{2}$

These two critical values divide the real number line into three distinct test intervals:
Interval I: $x<-5$
Interval II: $-5<x<\frac{3}{2}$
Interval III: $x>\frac{3}{2}$

We will now test a value from each interval to determine where the inequality holds true:

Testing Interval I ($x<-5$):
Let us choose $x=-6$.
The left-hand side (LHS) becomes:
$\frac{1}{-6+5}=\frac{1}{-1}=-1$
The right-hand side (RHS) becomes:
$\frac{1}{2(-6)-3}=\frac{1}{-12-3}=-\frac{1}{15}$
Since $-1\le -\frac{1}{15}$ is a true statement, the inequality holds on the interval $x<-5$.

Testing Interval II ($-5<x<\frac{3}{2}$):
Let us choose $x=0$.
The left-hand side (LHS) becomes:
$\frac{1}{0+5}=\frac{1}{5}$
The right-hand side (RHS) becomes:
$\frac{1}{2\left(0\right)-3}=-\frac{1}{3}$
Since $\frac{1}{5}\le -\frac{1}{3}$ is false (a positive number cannot be less than a negative number), the inequality does not hold on this interval.

Testing Interval III ($x>\frac{3}{2}$):
This interval represents the region where the denominators are positive. When comparing the reciprocals, the inequality is satisfied, establishing the validity of this region as part of the solution set.

Combining the valid regions, we obtain the solution:
$x<-5$ or $x>\frac{3}{2}$