Chapter 8.6, Problem 7E

To determine

To calculate: The given inequality and check solution.

Answer to Problem 7E

the solution is $x<0$ or $\frac{2}{3}<x<1$ .

Explanation of Solution

Given information: Inequality $\frac{1}{x-1}+\frac{2}{x}<0$ .

Calculation:

Inequality $\frac{1}{x-1}+\frac{2}{x}<0$ .

The excluded value for this inequality is $0$ and $1$ .

Solve the related equation.

  $\frac{1}{x-1}+\frac{2}{x}=0$

The LCD for the terms is $x\times \left(x-1\right)$ .

Multiply both side by LCD

  $\begin{array}{l}x+2\times \left(x-1\right)=0\\ 3x-2=0\\ x=\frac{2}{3}\end{array}$

Draw vertical lines at the excluded value and at the solutions to separates the number line into intervals.

  

Now test a sample value in each interval to determine whether the values in the interval satisfy the inequality.

Test $x=-1$

  $\begin{array}{l}\frac{1}{-1-1}+\frac{2}{-1}<0\\ -\frac{1}{2}-\frac{2}{1}<0\\ -\frac{5}{2}<0\end{array}$

Test $x=\frac{1}{2}$

  $\begin{array}{l}\frac{2}{-1}+\frac{4}{1}<0\\ 2<0\times \end{array}$

Test $x=\frac{8}{10}$

  $\begin{array}{l}\frac{10}{-2}+\frac{5}{2}<0\\ -\frac{5}{2}<0\end{array}$

Test $x=2$

  $\begin{array}{l}1+1<0\\ 2<0\times \end{array}$

The statement is true for $x=-1$ and $x=\frac{8}{10}$ . Therefore, the solution is $x<0$ or $\frac{2}{3}<x<1$ .