The set of values of $x$ satisfying the system of inequations $5 x + 2 < 3 x + 8$ and $\frac{x + 2}{x - 1} < 4$ is

(- \infty, 1)

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(2, 3)

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(- \infty, 3)

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

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Solution

The correct option is D $(- \infty, 1) \cup (2, 3)$
$5 x + 2 < 3 x + 8$
$= 2 x < 6$
$= x < 3$
$x \in (- \propto, 3) . . . . . . . . . . . . . . . (i)$
$\frac{x + 2}{x - 1} < 4$
$= \frac{x + 2}{x - 1} - 4 < 0$
$= \frac{x + 2 - 8 x + 4}{x - - 1} < 0$
$= \frac{- 3 x + 6}{x - 1} < 0$
$c a s e 1$
$- 3 x + 6 < 0 o r x - 1 > 0$
$= x > 2 x > 1$
$c a s e 2$
$- 3 x + 6 > 0 o r x - 1 < 0$
$= x < 2 x < 1$
$H e n c e x \in (- \propto, 1) \cup (2, \propto) . . . . . . . . . . . . (i i)$
$f r o m int e r sec t i o n o f (i) a n d (i i)$
$x \in (- \propto, 1) \cup (2, 3)$

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