In the following exercises, solve each rational inequality and write the solution in interval notation.

$\frac{6x}{x-6}>2$

In the following exercises, we have to solve rational inequality and write the solution in interval notation.

$\frac{6x}{x-6}>2$

The given function is$\frac{6x}{x-6}>2$

Subtract by $2$on both side

localid="1645253020043" $$\begin{gathered} \frac{6x}{x-6}-2>2-2 \\ \frac{6x}{x-6}-2>0 \end{gathered}$$

Simplifying the function

$$\begin{gathered} \frac{6x}{x-6}-\frac{2(x-6)}{x-6}>0 \\ \frac{6x-2(x-6)}{x-6}>0 \\ \frac{6x-2x+12}{x-6}>0 \\ \frac{4x+12}{x-6}>0 \end{gathered}$$

The rational expression will be zero when the numerator is zero. Since $4x+12=0$when $x=-3$, then $-3$is a critical point.

The rational expression will be undefined when the denominator is zero. Since $x-6=0$when $x=6$, then $6$is a critical point.

The critical points are $-3$and $6$.

The number line is divided into three intervals:

$(-\infty ,-3)(-3,6)(6,\infty )$

To find the sign of each factor in an interval, we choose any point in that interval and use it as a test point. Any point in the interval will give the expression the same sign, so we can choose any point in the interval.

$\mathit{I}\mathit{n}\mathit{t}\mathit{e}\mathit{r}\mathit{v}\mathit{a}\mathit{l}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{)}$

The number $-4$is in the interval $(-\infty ,-3)$. Test $x=-4$in the expression in the numerator and the denominator.

$\frac{4x+12}{x-6}=\frac{4(-4)+12}{-4-6}=\frac{-16+12}{-10}=\frac{-4}{-10}=\frac{2}{5}$

The number $1$is in the interval $(-3,6)$. Test $x=1$in the expression in the numerator and the denominator.

$\frac{4x+12}{x-6}=\frac{4\times 1+12}{1-6}=\frac{4+12}{-5}=-\frac{16}{5}$

The quotient is marked negative in the interval$(-3,6)$.

The number $7$is in the interval $(6,\infty )$. Test $x=7$in the expression in the numerator and the denominator.

$\frac{4x+12}{x-6}=\frac{4\times 7+12}{7-6}=\frac{28+12}{1}=40$

The quotient is marked positive in the interval $(6,\infty )$.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals$(-\infty ,-3)$and$(6,\infty )$are solutions.

The solution in interval notation is$(-\infty ,-3)\cup (6,\infty )$