All the values of x satisfying the inequality $\frac{1}{x+5} \leq \frac{1}{2 x-3}$ are

We start with: $\frac{1}{x+5} \leq \frac{1}{2x-3}$

When dealing with inequalities involving fractions, the key insight is that we can't simply "cross multiply" like we do with equations because we don't know if we're multiplying by positive or negative numbers. Instead, we move everything to one side to analyze the sign of the expression.

$\frac{1}{x+5} - \frac{1}{2x-3} \leq 0$

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To subtract these fractions, we need a common denominator: $(x+5)(2x-3)$

$\frac{1}{x+5} - \frac{1}{2x-3} = \frac{(2x-3) - (x+5)}{(x+5)(2x-3)}$

Let's simplify the numerator carefully:

$(2x-3) - (x+5) = 2x - 3 - x - 5 = x - 8$

So our inequality becomes: $\frac{x-8}{(x+5)(2x-3)} \leq 0$

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Critical points are values where our expression either equals zero or becomes undefined. These points divide the number line into intervals where the expression maintains a consistent sign.

Where the expression equals zero (numerator = 0):

$x - 8 = 0 \rightarrow x = 8$

Where the expression is undefined (denominator = 0):

$x + 5 = 0 \rightarrow x = -5$

$2x - 3 = 0 \rightarrow x = \frac{3}{2}$

Important: We must exclude $x = -5$ and $x = \frac{3}{2}$ from our final answer because division by zero is undefined.

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Our critical points $x = -5$, $\frac{3}{2}$, $8$ divide the number line into four intervals. Let's analyze the sign of each factor:

Factor Analysis:

$(x-8)$: negative when $x < 8$, positive when $x > 8$

$(x+5)$: negative when $x < -5$, positive when $x > -5$

$(2x-3)$: negative when $x < \frac{3}{2}$, positive when $x > \frac{3}{2}$

Sign Table:

Interval	$(x-8)$	$(x+5)$	$(2x-3)$	$\frac{x-8}{(x+5)(2x-3)}$
$x < -5$	$(-)$	$(-)$	$(-)$	$\frac{(-)}{(-)(-))} = \frac{(-)}{(+)} = (-)$
$-5 < x < \frac{3}{2}$	$(-)$	$(+)$	$(-)$	$\frac{(-)}{(+)(-)} = \frac{(-)}{(-)} = (+)$
$\frac{3}{2} < x < 8$	$(-)$	$(+)$	$(+)$	$\frac{(-)}{(+)(+)} = \frac{(-)}{(+)} = (-)$
$x > 8$	$(+)$	$(+)$	$(+)$	$\frac{(+)}{(+)(+)} = \frac{(+)}{(+)} = (+)$

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We want $\frac{x-8}{(x+5)(2x-3)} \leq 0$, which means we need the expression to be negative or zero.

From our sign analysis:

Negative regions: $x < -5$ and $\frac{3}{2} < x < 8$

Zero at: $x = 8$ (included because of $\leq$)

Exclude: $x = -5$ and $x = \frac{3}{2}$ (undefined points)

Therefore: $x \in (-\infty, -5) \cup \left(\frac{3}{2}, 8\right]$