Question

Solve each inequality algebraically.

$\frac{1}{x-2}<\frac{2}{3x-9}$

Short answer

Solution to the inequality $\frac{1}{x-2}<\frac{2}{3x-9}$is $(-\infty ,2)\cup (3,5)$.

Step-by-step solution

Step 1. Given information  

We have been given an inequality $\frac{1}{x-2}<\frac{2}{3x-9}$.

We have to solve this inequality algebraically.

Step 2. Write the inequality so that a polynomial or rational expression f is on the left side and zero is on the right side  

$$\begin{gathered} \frac{1}{x-2}<\frac{2}{3x-9} \\ \frac{1}{x-2}-\frac{2}{3x-9}<0 \\ \frac{(3x-9)-2(x-2)}{(x-2)(3x-9)}<0(\mathrm{Take}\mathrm{LCD}) \\ \frac{3x-9-2x+4}{(x-2)(3x-9)}<0 \\ \frac{x-5}{(x-2)(3x-9)}<0 \end{gathered}$$

Step 3. Determine the real numbers at which the expression f equals zero and at which the expression f is undefined. 

Assume $f\left(x\right)=\frac{x-5}{(x-2)(3x-9)}$.

$$\begin{gathered} x-5=0 \\ x=5 \\ x-2=0 \\ x=2 \\ 3x-9=0 \\ 3x=9 \\ x=3 \end{gathered}$$

Step 4. Form the intervals  

Using the values of x found in previous step, we can divide the real numbers in the intervals:

$(-\infty ,2)\cup (2,3)\cup (3,5)\cup (5,\infty )$

Step 5. Select a number in each interval and evaluate f at the number   

Create the following table:

Interval	$(-\infty ,2)$	$(2,3)$	$(3,5)$	$(5,\infty )$
Number chosen	0	2.5	4	6
Value of f	$f\left(0\right)=-\frac{5}{18}$	$f(2.5)=\frac{10}{3}$	$f\left(4\right)=-\frac{1}{6}$	$f\left(6\right)=\frac{1}{36}$
Conclusion	negative	positive	negative	positive

Step 6. Identify the interval 

Since we want to know where f is negative, we conclude that $f\left(x\right)<0$ in the interval $(-\infty ,2)\cup (3,5)$.