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

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

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Answer 1 · 2016-11-08T06:20:36

$x < - 3$ $x< -3$ or $x > 4$ $x> 4$

Explanation:

The numerator is positive for $x > 4$ $x >4$ whereas the denominator is positive for $x > - 3$ $x > -3$ . It comes that for $x < - 3$ $x< -3$ both of them are negative so that the whole fraction is positive. The same happens for $x > 4$ $x>4$ . On the other hand for $- 3 < x < 4$ $-3 < x< 4$ the fraction is negative. Consequently the fraction is poisitive
for $x < - 3$ $x< -3$ and for $x > 4$ $x>4$

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

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How do you solve (x-4)/(x+3)>0x−4x+3>0(x-4)/(x+3)>0?

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