How do you solve the inequality $5 + \frac{1}{x} > \frac{16}{x}$ $5+1/x>16/x$ ?

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Answer 1 · 2017-01-18T16:35:57

The answer is $x \in] - \infty, 0 [\cup] 3, + \infty [$ $x in ] -oo,0 [ uu ] 3, +oo[$

Let's rewrite the inequality

$5 + \frac{1}{x} - \frac{16}{x} > 0$ $5+1/x-16/x>0$

$5 - \frac{15}{x} > 0$ $5-15/x>0$

$\frac{5 (x - 3)}{x} > 0$ $(5(x-3))/x>0$

Let $f (x) = \frac{5 (x - 3)}{x}$ $f(x)=(5(x-3))/x$

The domain of $f (x)$ $f(x)$ is $D_f(x)=RR-{0}$

Now we can do a sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaaaa)$ $0$ $color(white)(aaaaaaa)$ $3$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x$ $color(white)(aaaaaaaa)$ $-$ $color(white)(aa)$ $∥$ $color(white)(aa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-3$ $color(white)(aaaaa)$ $-$ $color(white)(aa)$ $∥$ $color(white)(aa)$ $-$ $color(white)(aaaa)$ $+$

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

Therefore,

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

How do you solve the inequality 5+1/x>16/x5+1x>16x5+1/x>16/x?