How do solve $\frac{1}{x + 2} \geq \frac{1}{3}$ $1/(x+2)>=1/3$ algebraically?

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Answer 1 · 2016-09-12T07:26:49

$x \in (- 2, 1] .$ $x in (-2, 1].$ .

If #x>-2, x+2 is positive, Cross multiplication would not change the

in-between inequality sign. So,

$x + 2 \leq 3 \to x \leq 1$ $x+2<=3 to x <=1$ .

And so, #x in (-2, 1].

When $x < - 2, x + 2 < 0$ $x<-2, x+2<0$ . Now, cross multiplication reverses the

inequality sign. So,

$x + 2 \geq 3 \to x \geq 1$ $x+2>=3 to x>=1$ . As $x < - 2$ $x<-2$ , this is an extraneous solution..

It follows that the answer is , $x \in (- 2, 1]$ $x in(-2, 1]$ .

How do solve 1/(x+2)>=1/31x+2≥131/(x+2)>=1/3 algebraically?