How do you solve $(2t+7)/(t-4)>=3$ using a sign chart?

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Answer 1 · 2017-02-09T06:08:44

The solution is $t in ]4, 19]$

Rewrite the equation

$(2t+7)/(t-4)>=3$

$(2t+7)/(t-4)-3>=0$

$((2t+7)-3(t-4))/(t-4)>=0$

$(2t+7-3t+12)/(t-4)>=0$

$(-t+19)/(t-4)>=0$

Let $f(t)=(-t+19)/(t-4)$

We can construct the sign chart

$color(white)(aaaa)$ $t$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaaaaa)$ $4$ $color(white)(aaaaaaaa)$ $19$ $color(white)(aaaaaa)$ $+oo$

$color(white)(aaaa)$ $t-4$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $||$ $color(white)(aaaa)$ $+$ $color(white)(aaaaa)$ $+$

$color(white)(aaaa)$ $19-t$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $||$ $color(white)(aaaa)$ $+$ $color(white)(aaaaa)$ $-$

$color(white)(aaaa)$ $f(t)$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $||$ $color(white)(aaaa)$ $+$ $color(white)(aaaaa)$ $-$

Therefore,

$f(t)>=0$ when $t in ]4, 19]$

How do you solve (2t+7)/(t-4)>=3(2t+7)/(t-4)>=3 using a sign chart?