91Ó°ÊÓ

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

$\frac{3x}{x−5}<1$

The function is $\frac{3x}{x−5}<1$

Subtract by $1$on both side

$$\begin{gathered} \frac{3x}{x−5}-1<1-1 \\ \frac{3x}{x−5}-1<0 \end{gathered}$$

Now simplifying the function

$$\begin{gathered} \frac{3x}{x−5}-1·\frac{x−5}{x−5}<0 \\ \frac{3x}{x−5}-\frac{x−5}{x−5}<0 \\ \frac{3x-(x−5)}{x−5}<0 \\ \frac{3x-x+5}{x−5}<0 \\ \frac{2x+5}{x−5}<0 \end{gathered}$$

The rational expression will be zero when the numerator is zero. Since $2x+5=0$when $x=-\frac{5}{2}$, then $-\frac{5}{2}$is a critical point.

The rational expression will be undefined when the denominator is zero. Since$x-5=0$when$x=5$, then$5$is a critical point.

The critical points are $-\frac{5}{2}$and $5$.

The number line is divided into three intervals:$(-∞,-\frac{5}{2})(-\frac{5}{2},5)(5,∞)$

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.

$\mathit{I}\mathit{n}\mathit{t}\mathit{e}\mathit{r}\mathit{v}\mathit{a}\mathit{l}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{∞}\mathbf{,}\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}\mathbf{)}$

The number $-3$is in the interval $(-∞,-\frac{5}{2})$. Test $x=-3$in the expression in the numerator and the denominator.

$\frac{2x+5}{x−5}=\frac{2(-3)+5}{-3−5}=\frac{-6+5}{-8}=\frac{1}{8}$

The quotient positive in the interval $(-∞,-\frac{5}{2})$.

The number $1$is in the interval $(-\frac{5}{2},5)$. Test $x=1$in the expression in the numerator and the denominator.

$\frac{2x+5}{x−5}=\frac{2×1+5}{1−5}=\frac{2+5}{-4}=\frac{7}{-4}$

The quotient is marked negative in the interval $(-\frac{5}{2},5)$.

The number $6$is in the interval $(5,∞)$. Test $x=6$in the expression in the numerator and the denominator.

$\frac{2x+5}{x−5}=\frac{2×6+5}{6−5}=\frac{12+5}{1}=17$

The quotient is marked negative in the interval $(5,∞)$.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals$(-∞,-\frac{5}{2})$and$(5,∞)$are solutions.

The solution in interval notation is$(-∞,-\frac{5}{2})∪(5,∞)$