How do you solve $4/(x-3)<2$ using a sign chart?

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Answer 1 · 2017-02-15T13:03:01

The answer is $x in ]-oo, 3[uu]5, +oo[$

Let's rearrange the equation

$4/(x-3)-2<0$

$(4-2(x-3))/(x-3)<0$

$(4-2x+6)/(x-3)<0$

$(10-2x)/(x-3)<0$

Let $f(x)=(10-2x)/(x-3)$

The domain of $f(x)$ is $D_f(x)=RR-{3}$

Now, we build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaaa)$ $-oo$ $color(white)(aaaaa)$ $3$ $color(white)(aaaaaaa)$ $5$ $color(white)(aaaaaa)$ $+oo$

$color(white)(aaaa)$ $x-3$ $color(white)(aaaaa)$ $-$ $color(white)(aaa)$ $||$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $10-2x$ $color(white)(aaaa)$ $+$ $color(white)(aa)$ $||$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $-$

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

Therefore,

$f(x)<0$ when $x in ]-oo, 3[uu]5, +oo[$

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