Question
Solve the following inequalities and graph their solution set :
$\frac{x+8}{x+1}>1$.
$\frac{x+8}{x+1}>1$.
✓
Answer
The inequality $\frac{x+8}{x+1}$ is equivalent to
$\frac{x+8}{x+1}-1>0 \Leftrightarrow \frac{x+8-x-1}{x+1}>0 $
$ \Rightarrow \frac{7}{x+1}$
$= 0$
$\text { but } \frac{a}{b}>0, a>0 \Leftrightarrow b>$
Thus, $\frac{7}{x+1}>0,7>0$
$\Rightarrow x + 1 > 0$ or $x > -1$
The graph of this set is
$\frac{x+8}{x+1}-1>0 \Leftrightarrow \frac{x+8-x-1}{x+1}>0 $
$ \Rightarrow \frac{7}{x+1}$
$= 0$
$\text { but } \frac{a}{b}>0, a>0 \Leftrightarrow b>$
Thus, $\frac{7}{x+1}>0,7>0$
$\Rightarrow x + 1 > 0$ or $x > -1$
The graph of this set is
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