Henry tried to write the set $M=\left\{x\in\mathbb{R}∶|-4-2x|\leq3\right\}$ as an interval. In which step of his solution did Henry make a mistake?
Henry´s solution:
(1) Henry figured that the interval he was looking for was the solution to the inequality: $$|-4-2x|\leq3$$ (2) First, he modified the expression inside the absolute value: $$|4+2x|\leq3$$ (3) Subsequently, he factored out the number $2$: $$2|x+2|\leq3$$ (4) Then, he divided the inequality by $2$: $$|x+2|\leq1.5$$
(5) Finally, he realized that the set $M$ contains all real numbers whose distance from number $2$ is less than or equal to $1.5$, i.e., $$M=[2-1.5;2+1.5]=[0.5;3.5]$$
The mistake is in step (2). The correct simplification should be: $$|-4-2x|=-|4+2x|$$
The mistake is in step (3). It is not possible to factor out a number from an expression inside the absolute value.
The mistake is in step (4). If we divide an inequality by the number $2$, we have to change the sign. Therefore, we should have got: $$|x+2|\geq1.5$$
The mistake is in step (5). The set $M$ contains all real numbers whose distance from number $-2$ is less than or equal to $1.5$, i.e., $$M=[-2-1.5;-2+1.5]=[-3.5;-0.5]$$