Question
Solve the following linear inequations in R: $\frac{1}{\text{x}-1}\leq2$

Answer

$\frac{1}{\text{x}-1}\leq2$ $\frac{1}{\text{x}-1}-2\leq0$ $\frac{1-2(\text{x}-1)}{\text{x}-1}\leq0$ $\frac{1-2\text{x}+2}{\text{x}-1}\leq0$ $\frac{3-2\text{x}}{\text{x}-1}\leq0$ Case 1: $3-2\text{x}\geq0$ and $\text{x}-1<0$ $\Rightarrow\text{x}\leq\frac{3}{2}$ and $\text{x}<1$ Case 2: $3-2\text{x}\leq0$ and $\text{x}-1>0$ $\Rightarrow\text{x}\geq\frac{3}{2}$ and $\text{x}>1$ Hence the solution set is $(-\infty,1)\cup\Big[\frac{3}{2},\infty\Big)$

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