F or m u l a f or ab so l u t e v a l u e ∣ a ∣ = a , f or a ≥ 0 or ∣ a ∣ = − a , f or a < 0 − 3∣2 x + 6∣ = − 12 − 3 ( 2 x + 6 ) = − 12 or − 3 ( − 2 x − 6 ) = − 12 − 6 x − 18 = − 12 or 6 x + 18 = − 12 − 6 x = 6 or 6 x = − 30 x = − 1 or x = − 5 S o l u t i o n s a re − 1 an d − 5.
6x+18=-12 6x=-12 + -18 6x=30 x=5
To solve the equation − 3∣2 x + 6∣ = − 12 , we simplify to ∣2 x + 6∣ = 4 and consider two cases. The solutions are x = − 1 and x = − 5 , both of which can be verified by substituting back into the original equation. Thus, the final answers are x = − 1 and x = − 5 .
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