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

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

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Answer 1 · 2016-08-07T20:16:57

$x \in (- 4, - 3)$ $x in (-4, -3)$

Explanation:

There are three intervals within which the sign of the quotient does not change:

$(- \infty, - 4)$ $(-oo, -4)$

$(- 4, - 3)$ $(-4, -3)$

$(- 3, \infty)$ $(-3, oo)$

We do not need to consider the points $x = - 4$ $x=-4$ and $x = - 3$ $x=-3$ as the former gives an undefined quotient and the latter a zero quotient. So they are not in the solution set.

$color(white)()$
Case $bb(x in (-oo, -4))$

Both $(x+3) < 0$ and $(x+4) < 0$ , so the quotient is positive.

So this interval is not part of the solution set.

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Case $bb(x in (-4, -3))$

In this interval, $(x+3) < 0$ but $(x+4) > 0$ , so the quotient is negative.

So this interval is part of the solution set.

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Case $bb(x in (-3, oo))$

In this interval, $(x+3) > 0$ and $(x+4) > 0$ , so the quotient is positive.

So this interval is not part of the solution set.

$color(white)()$
Conclusion

The solution set is the interval $(-4, -3)$

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

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