How do you solve $- 3 x + 2 > 11$ $-3x + 2 >11$ or $5 (x - 2) < 0$ $5(x - 2) <0$ ?

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Answer 1 · 2017-07-13T10:53:08

Solution : $x < - 3$ $x < -3$ , in interval notation : $(- \infty, - 3)$ $(-oo, -3)$

1) $- 3 x + 2 > 11 or - 3 x > 9 or - x > 3 or x < - 3$ $-3x +2 > 11 or -3x > 9 or -x > 3 or x < -3$

2) $5 (x - 2) < 0 or x - 2 < 0 or x < 2$ $5 (x-2) < 0 or x-2 < 0 or x < 2$

$x < - 3$ $x < -3$ satisfies both inequaity.

Solution : $x < - 3$ $x < -3$ , in interval notation : $(- \infty, - 3)$ $(-oo, -3)$ [Ans]

How do you solve −3x+2>11-3x + 2 >11 or 5(x−2)<05(x - 2) <0?