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

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

Accepted Answer

x &lt; -5 or \cfrac{3}{2} &lt; x \leq 8. There are two critical points for the inequality to consider: x=-5 and x=3/2 Region I: x is greater than 3/2 In this scenario, both the terms would be positive; cross-multiplying, we get the relation 2 x − 3 ≤ x + 5 2x-3\le x+5 2 x − 3 ≤ x + 5 Giving the boundary x ≤ 8 x\le8 x ≤ 8 , hence giving us

All the values of x satisfying the inequality 1x+5≤12x−3\cfrac{1}{x + 5} \leq \cfrac{1}{2x - 3}x+51​≤2x−31​ are

All the values of x satisfying the inequality 1x+5≤12x−3\cfrac{1}{x + 5} \leq \cfrac{1}{2x - 3}x+51​≤2x−31​ are

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

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