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

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

Accepted Answer

We are given the inequality: $$ \frac{1}{x + 5} \leq \frac{1}{2x - 3} $$ --- #### **Step 1: Domain restrictions** * $x + 5 eq 0 \Rightarrow x eq -5$ * $2x - 3 eq 0 \Rightarrow x eq \frac{3}{2}$ So, the expression is **undefined at** $x = -5$ and $x = \frac{3}{2}$ --- #### **Step 2: Bring to one side** $$ \frac{1}{x + 5} - \frac{1}{2x - 3} \leq 0 $$ Take LCM and simplify: $$ \frac{(2x - 3) - (x + 5)}{(x + 5)(2x - 3)} \leq 0 \Rightarrow \frac{x - 8}{(x + 5)(2x - 3)} \leq 0 $$ --- #### **Step 3: Sign Analysis** Find the critical points where the expression is zero or undefined: * Numerator zero at $x = 8$ * Denominator zero at $x = -5$ and $x = \frac{3}{2}$ These divide the number line into intervals: 1. $(-\infty, -5)$ 2. $(-5, \frac{3}{2})$ 3. $(\frac{3}{2}, 8)$ 4. $(8, \infty)$ Test sign in each: * **Interval 1: $x = -6$** $\frac{-14}{(-1)(-15)} = \frac{-14}{15} < 0$ * **Interval 2: $x = 0$** $\frac{-8}{5 \cdot (-3)} = \frac{-8}{-15} > 0$ * **Interval 3: $x = 2$** $\frac{-6}{7} < 0$ * **Interval 4: $x = 10$** $\frac{2}{15 \cdot 17} > 0$ So the expression is **≤ 0** in: * $(-\infty, -5)$ * $\left( \frac{3}{2}, 8 ight]$ Note: Exclude $x = -5$ and $x = \frac{3}{2}$ since they make the expression undefined. --- ### ✅ Final Answer: $$ x \in (-\infty, -5) \cup \left( \frac{3}{2}, 8 ight] $$ This matches **Option A**. --- ### **Correct Option: A**

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