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The set of values of x which satisfy 5x+2 < 3x+8 and $\frac{(x + 2)}{x - 1} < 4$ is

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$(- \infty, 1) \cup (2, 3)$

(2, 3)

(1, 3)

$(- \infty, 1)$

answer is B.

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Detailed Solution

$\begin{array}{l} 2 x < 6 and \frac{x + 2 - 4 (x - 1)}{x - 1} < 0 \\ \frac{3 x - 6}{x - 1} > 0 x \in (- \infty, 1) \cup (2, \infty) \end{array}$

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