How do you solve $| 3 x + 2 | - 1 \geq 10$ $abs(3x + 2) - 1 >=10$ ?

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How do you solve $| 3 x + 2 | - 1 \geq 10$ $abs(3x + 2) - 1 >=10$ ?

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Answer 1 · 2016-07-11T03:19:26

$x \leq - \frac{13}{3}$ $x <= -13/3$
$x \geq 3$ $x >= 3$

Explanation:

$I 3 x + 2 I \geq 11$ $I3x + 2I >= 11$
Separate solving into 2 parts:
a. $(3 x + 2) \geq 11$ $(3x + 2) >= 11$
$3 x \geq 9 \circ$ $3x >= 9@$ x >= 3 $b .$ $b.$ -(3x + 2) >= 11 - 3x - 2 >= 11 - 3x >= 13x <= -13/3 $N o t e . T h e 2 e n d p \oint s$ $Note. The 2 end points$ (-13/3)# and (3) are included in the solution set.
Graph:

=============== -13/3 ---------0---------- 3 =============

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How do you solve $| 3 x + 2 | - 1 \geq 10$ $abs(3x + 2) - 1 >=10$ ?

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How do you solve abs(3x + 2) - 1 >=10|3x+2|−1≥10abs(3x + 2) - 1 >=10?

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How do you solve $| 3 x + 2 | - 1 \geq 10$ $abs(3x + 2) - 1 >=10$ ?