How do you solve $| 2 x + 9 | \leq 3$ $abs(2x+9) ≤ 3$ ?

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Answer 1 · 2017-02-28T23:11:54

$- 6 \leq x \leq - 3$ $-6 le x le-3$

Because it has the absolute value sign, you can assume there are two equations:
$2 x + 9 \leq 3$ $2x+9le3$
and
$- (2 x + 9) \leq 3$ $-(2x+9)le3$

Lets solve for the first one.
$2 x \leq - 6$ $2xle-6$
$x \leq - 3$ $xle-3$

Now onto the second one.
$2 x + 9 \geq - 3$ $2x+9ge-3$
$2 x \geq - 12$ $2xge-12$
$x \geq - 6$ $xge-6$

Now combine the two answers.
$x \leq - 3$ $xle-3$ and $x \geq - 6$ $xge-6$ for $- 6 \leq x \leq - 3$ $-6lexle-3$

How do you solve |2x+9|≤3abs(2x+9) ≤ 3?