# How do you solve (x+7)/(x-4)<0?

Aug 8, 2017

$- 7 < x < 4$

#### Explanation:

For $\frac{x + 7}{x - 4}$ to be less than $0$, the top part must be less than $0$, or negative.

An $x$ value of $- 7$ will cancel the top part, but only give us $0$, anything less than $- 7$ will give a negative fraction, and therefore less than $0$.

Proof:
$\frac{x + 7}{x - 4} < 0$

$\frac{\left(x + 7\right) \cancel{\left(x - 4\right)}}{\cancel{\left(x - 4\right)}} < 0 \left(x - 4\right)$

$x + 7 < 0$

$x + 7 - 7 < - 7$

$x < - 7$

However, if top and bottom are negative, it will be positive, and any $x$ value less than $- 7$ will give a positive value. However if $x < 4$, then the bottom value will be negative, giving a negative value.

$- 7 < x < 4$

Aug 8, 2017

The solution is $x \in \left(- 7 , 4\right)$

#### Explanation:

Let $f \left(x\right) = \frac{x + 7}{x - 4}$

We can build the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 7$$\textcolor{w h i t e}{a a a a a a}$$4$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 7$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 4$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(x\right) < 0$ when $x \in \left(- 7 , 4\right)$

graph{(x+7)/(x-4) [-41.1, 41.14, -20.54, 20.55]}

Aug 8, 2017

$x \in \left(- 7 , 4\right)$

#### Explanation:

Given:

$\frac{x + 7}{x - 4} < 0$

Note that since the linear expressions $\left(x + 7\right)$ and $\left(x - 4\right)$ each occur once, the rational expression will change sign at the points $x = - 7$ and $x = 4$. It has a vertical asymptote at $x = 4$ and intercepts the $x$ axis at $x = - 7$.

For large positive or negative values of $x$, the rational expression is positive, so the interval in which it is negative is precisely $\left(- 7 , 4\right)$

graph{(y-(x+7)/(x-4))(x-3.99+y*0.0001) = 0 [-19.55, 20.45, -10.12, 9.88]}

Aug 8, 2017

The answer is x in (-7;4). See explanation.

#### Explanation:

First we have to calculate the domain of the rational expression. As the denominator cannot be zero, the excluded values are:

$x - 4 \ne 0 \implies x \ne 4$

Now we can solve the inequality.

## $\frac{x + 7}{x - 4} < 0$

We can change the rational inequality to quadratic inequality by multiplying it by the square of the denominator:

## $\left(x + 7\right) \left(x - 4\right) < 0$

If we graph the quadratic function:

graph{(x-4)*(x+7) [-36.52, 36.52, -18.22, 18.35]}

we see that it takes negative values for x in (-7;4), so this interval is also the solution of the initial rational inequality