How do you solve $x/(x+3)>=2$ ?

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Answer 1 · 2016-12-27T14:28:09

The answer is $x in [-6, -3[$

We cannot do crossing over.

We rewrite the inequality as $x/(x+3)-2>=0$

$(x-2(x+3))/(x+3)>=0$

$(x-2x-6)/(x+)=(-x-6)/(x+3)>=0$

Let $f(x)=-(x+6)/(x+3)$

We can make a time chart

$color(white)(aaaa)$ $x$ $color(white)(aaaaa)$ $-oo$ $color(white)(aaaa)$ $-6$ $color(white)(aaaaa)$ $-3$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $-x-6$ $color(white)(aaaa)$ $+$ $color(white)(aaaaaa)$ $-$ $color(white)(aa)$ $∣∣$ $color(white)(aa)$ $-$

$color(white)(aaaa)$ $x+3$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaaaa)$ $-$ $color(white)(aa)$ $∣∣$ $color(white)(aa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaa)$ $-$ $color(white)(aaaaaa)$ $+$ $color(white)(aa)$ $∣∣$ $color(white)(aa)$ $-$

Therefore,

$f(x)>=0$ , when $x in [-6, -3[$

How do you solve x/(x+3)>=2x/(x+3)>=2?