# How do you solve (x+68)/(x+8)>=5?

May 7, 2017

$x \in \left(- 8 , 7\right]$

#### Explanation:

Given:

$\frac{x + 68}{x + 8} \ge 5$

Subtract $\frac{x + 68}{x + 8}$ from both sides to get:

$0 \ge 5 - \frac{x + 68}{x + 8} = \frac{5 \left(x + 8\right) - \left(x + 68\right)}{x + 8} = \frac{4 x - 28}{x + 8} = \frac{4 \left(x - 7\right)}{x + 8}$

Note that the right hand side is a function which is continuous and non-zero except at $x = - 8$ and $x = 7$. The function changes sign at each of these two points.

When $x = - 8$ the denominator is $0$ so the right hand side is undefined. So $- 8$ is not part of the solution set.

When $x = 7$ the numerator is $0$ and the inequality is satisfied. So $7$ is part of the solution set.

When $x < - 8$ or $x > 7$ then the signs of the numerator and denominator are the same, so the quotient is positive and the inequality is not satisfied.

When $x \in \left(- 8 , 7\right)$ the numerator is negative, the denominator is positive and the quotient is negative. So the inequality is satisfied.

So the solution set is:

$x \in \left(- 8 , 7\right]$

May 7, 2017

The solution is $x \in \left(- 8 , 7\right]$

#### Explanation:

We cannot do crossing over.

So, we simplify the inequality

$\frac{x + 68}{x + 8} \ge 5$

$\frac{x + 68}{x + 8} - 5 \ge 0$

$\frac{\left(x + 68\right) - 5 \left(x + 8\right)}{x + 8} \ge 0$

$\frac{x + 68 - 5 x - 40}{x + 8} \ge 0$

$\frac{28 - 4 x}{x + 8} \ge 0$

$\frac{4 \left(7 - x\right)}{x + 8} \ge 0$

Let $f \left(x\right) = \frac{4 \left(7 - x\right)}{x + 8}$

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}$$- 8$$\textcolor{w h i t e}{a a a a a a a a}$$7$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 8$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

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

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(- 8 , 7\right]$