How do you solve compound inequalities $8 - 4x < -3x + 5$ or $x + 8 > 4 + 3x$ ?

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Answer 1 · 2015-05-09T03:59:26

(1) x + 8 > 4 + 3x -> 2x < 8 - 4 -> x < 2

(2) 8 - 4x < -3x + 5 -> 8 - 5 < 4x - 3x -> x > 3
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The solution set are the 2 open intervals: (-infinity , 2) and (3, +infinity)

How do you solve compound inequalities 8 - 4x < -3x + 58 - 4x < -3x + 5 or x + 8 > 4 + 3xx + 8 > 4 + 3x?