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

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Answer 1 · 2016-07-21T06:21:22

$x \in (- \infty, - 4) \cup (0, \infty)$ $x in (-oo, -4) uu (0, oo)$

If $x = 0$ $x = 0$ then the denominator is $0$ $0$ so the quotient is undefined. So $x = 0$ $x = 0$ is not part of the solution set.

That leaves two cases:

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Case $x > 0$ $bb(x > 0)$

Multiply both sides of the inequality by $x$ $x$ to get:

$x + 4 > 0$ $x+4 > 0$

Subtract $4$ $4$ from both sides to get:

$x > - 4$ $x > -4$

Since $x > 0$ $x > 0$ , this holds already, so the whole of $(0, \infty)$ $(0, oo)$ is part of the solution set.

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Case $x < 0$ $bb(x < 0)$

Multiply both sides of the inequality by $0$ $0$ and reverse the inequality (since the multiplier is negative) to get:

$x + 4 < 0$ $x+4 < 0$

Subtract $4$ $4$ from both sides to get:

$x < - 4$ $x < -4$

So $(- \infty, - 4)$ $(-oo, -4)$ is part of the solution set.

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Conclusion

The solution set is the union of these two cases, namely:

$(- \infty, - 4) \cup (0, \infty)$ $(-oo, -4) uu (0, oo)$

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