How do you solve the inequality $7/(y+1)>7$ ?

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Answer 1 · 2016-12-17T14:48:39

The answer is $x in ] -1,0 [$

We rewrite the equation as

$7-7/(y+1)<0$

$(7(y+1)-7)/(y+1)<0$

$(7y+7-7)/(y+1)<0$

$(7y)/(y+1)<0$

Let $f(y)=(7y)/(y+1)$

and $y!=-1$

We do a sign chart

$color(white)(aaaa)$ $y$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-1$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $y$ $color(white)(aaaaaaaa)$ $-$ $color(white)(aa)$ $∥$ $color(white)(a)$ $-$ $color(white)(aa)$ $+$

$color(white)(aaaa)$ $y+1$ $color(white)(aaaa)$ $-$ $color(white)(aaa)$ $∥$ $color(white)(a)$ $+$ $color(white)(aa)$ $+$

$color(white)(aaaa)$ $f(y)$ $color(white)(aaaaa)$ $+$ $color(white)(aaa)$ $∥$ $color(white)(a)$ $-$ $color(white)(aa)$ $+$

Therefore,

$f(y)<0$ when $x in ] -1,0 [$

graph{7x/(x+1) [-41.1, 41.1, -20.55, 20.57]}

How do you solve the inequality 7/(y+1)>77/(y+1)>7?