|2x + 5| ≤ |x + 3| I have the answer listed in front of me but it's not helping me figure out how to get there.
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Welcome to Mathematics Stack Exchange. You could consider cases $x<-3$, $-3\le x\lt-2.5$, and $-2.5\le x$ separately, so you know the absolute values – J. W. Tanner Sep 22 '20 at 02:56
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I think I might have not asked for the right thing. The answer is $$[- \frac{8}{3},-2]$$ – timyorgut Sep 22 '20 at 03:20
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One way to solve this is to square both sides:
$$(2x+5)^2\le (x+3)^2$$
$$4x^2+20x+25\le x^2+6x+9$$
$$3x^2+14x+16\le 0$$
$$(3x+8)(x+2)\le0.$$
If a product of two terms is negative, one is negative and one is positive.
Can you take it from here?
J. W. Tanner
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I think I might have not asked for the right thing. The answer is $$[-\frac{8}{3},-2]$$ – timyorgut Sep 22 '20 at 03:30
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That looks right. In that interval, $3x+8$ is non-negative and $x+2$ is non-positive – J. W. Tanner Sep 22 '20 at 03:35