Using symbol lab I put this in $|x+4|\le |2x+10|$ and the answer I get is $x \le -6$ or $x\ge -14/3$, but when I manually worked out it was $\;x \ge -6\;$ or $\;x\ge -14/3$. My working out is in the description:The image of my working out
The simplest way to solve this inequation is to use the fact that, for any $A$, $|A|^2=A^2$ (this removes the absolute values), and that function $x^2$ is increasing for non-negative $x$. Thus $$|x+4|\le |2x+10|\iff(x+4)^2\le (2x+10)^2\iff3x^2+32x+84\ge 0.$$ Can you proceed?
You must consider the following cases: $$x\geq -4$$ and $$x\geq -5$$ or $$x\geq -4$$ and $$x<-5$$ or$$x<-4$$ and $$x<-5$$ If $$x\geq -4$$ and $$x\geq -5$$ then we have to solve $$x+4\le 2(x+5)$$ Can you proceed? The second case is not possible so you have to solve for $$-5\le x<-4$$: $$-x-4\le 2(x+5)$$ and for $$x<-5$$ you have to solve $$-x-4\le 2(-x-5)$$
Here you have a simple difference:
|𝑥+4|≤|2𝑥+10| <=> (𝑥+4)^2 ≤ (2𝑥+10)^2, as both sides are positive.
On the other hand, if 𝑥+4 ≤ 2𝑥+10 it is not necessarily true that it is equivalent to (𝑥+4)^2 ≤ (2𝑥+10)^2, as x+4 and 2x+10 can have any sign.
For example, -2 ≤ 1 is not equivalent to 4 ≤ 1.
This is the reason that you get different answers, as you are changing the condition you are giving the set of numbers that satisfy the inequality.