How do you solve $\frac{x + 7}{x - 3} > 0$ $(x+7)/(x-3)>0$ ?

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How do you solve $\frac{x + 7}{x - 3} > 0$ $(x+7)/(x-3)>0$ ?

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Answer 1 · 2017-02-06T10:58:04

The answer is $x \in] - \infty, - 7 [\cup] 3, + \infty [$ $x in ]-oo, -7 [uu] 3, +oo[$

Explanation:

Let $f (x) = \frac{x + 7}{x - 3}$ $f(x)=(x+7)/(x-3)$

Let's build the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a$ $color(white)(aaaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 7$ $-7$ $a a a a$ $color(white)(aaaa)$ $3$ $3$ $a a a a a$ $color(white)(aaaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 7$ $x+7$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 3$ $x-3$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a a$ $color(white)(aaaaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (x) > 0$ $f(x)>0$ when $x \in] - \infty, - 7 [\cup] 3, + \infty [$ $x in ]-oo, -7 [uu] 3, +oo[$

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How do you solve $\frac{x + 7}{x - 3} > 0$ $(x+7)/(x-3)>0$ ?

1 Answer

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