Solve the system of inequality:
$2 (3 x - 1) < 3 (4 x + 1) + 16, 4 (2 + x) < 3 x + 8$

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Solution

$(1) 2 (3 x - 1) < 3 (4 x + 1) + 16$
$6 x - 2 < 12 x + 3 + 16$
$6 x > - 5 - 16$
$x > \frac{- 21}{6}$
$x ε (\frac{- 21}{6}, \infty)$
$(2) 4 (2 + x) < 3 x + 8$
$8 + 4 x < 3 x + 8$
$x < 0$
$x ε (- \infty, 0)$
$x ε (\frac{- 21}{6}, 0)$

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