How do you solve $\frac{3 x + 2}{x - 4} < 0$ $(3x+2)/(x-4)<0$ ?

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How do you solve $\frac{3 x + 2}{x - 4} < 0$ $(3x+2)/(x-4)<0$ ?

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Answer 1 · 2017-06-06T11:12:36

The solution is $x \in (- \frac{2}{3}, 4)$ $x in (-2/3,4)$

Explanation:

Let $f (x) = \frac{3 x + 2}{x - 4}$ $f(x)=(3x+2)/(x-4)$

We build a sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- \frac{2}{3}$ $-2/3$ $a a a a a a a$ $color(white)(aaaaaaa)$ $4$ $4$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $3 x + 2$ $3x+2$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a a$ $color(white)(aaaaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $∣ ∣$ $||$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 4$ $x-4$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $∣ ∣$ $||$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a a$ $color(white)(aaaaaaa)$ $+$ $+$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $∣ ∣$ $||$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (x) < 0$ $f(x)<0$ , when $x \in (- \frac{2}{3}, 4)$ $x in (-2/3,4)$

graph{(3x+2)/(x-4) [-28.86, 28.85, -14.43, 14.45]}

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How do you solve $\frac{3 x + 2}{x - 4} < 0$ $(3x+2)/(x-4)<0$ ?

1 Answer

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