Find the solution set for the nonlinear inequality $\frac { 1 } { x - 3 } - \frac { 2 } { x - 2 } 0$ .
A) $( - \infty , - 3 ) \cup ( 2,4 )$
B) $( - \infty , 2 ] \cup [ 3,4 ]$
C) $( 2,3 )$
D) $( - \infty , 2 )$
E) $( - \infty , 2 ) \cup ( 3,4 )$

Question

Find the solution set for the nonlinear inequality $\frac { 1 } { x - 3 } - \frac { 2 } { x - 2 } > 0$ .
A) $( - \infty , - 3 ) \cup ( 2,4 )$
B) $( - \infty , 2 ] \cup [ 3,4 ]$
C) $( 2,3 )$
D) $( - \infty , 2 )$
E) $( - \infty , 2 ) \cup ( 3,4 )$

Quizplus · Accepted Answer

We begin by finding the critical values of x where the expression is undefined or equal to 0. The expression is undefined when either the denominator of the first term, $x-3$, or the denominator of the second term, $x-2$, is equal to 0. Hence, the critical values of x are $x=2,3$. We test the intervals separated by these critical values using test points, picking values less than 2, between 2 and 3, between 3 and 4, and greater than 4. We test x = 0 and x = 1 to the left of 2, x = 2.5 between 2 and 3, and x = 5 to the right of 4. Plugging these values into the expression, we get: $x=0:$ $\frac{1}{-3}-\frac{2}{-2} < 0$ which is false. $x=1:$ $\frac{1}{-2}-\frac{2}{-1} > 0$ which is true. $x=2.5:$ $\frac{1}{-.5}-\frac{2}{.5} > 0$ which is true. $x=4:$ $\frac{1}{1}-\frac{2}{2} < 0$ which is false. Therefore, the solution set is $( - \infty , 2 ) \cup ( 3,4 )$. The answer is $\boxed{ extbf{(E)}}$.

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Find the Solution Set for the Nonlinear Inequality $\frac { 1 } { x - 3 } - \frac { 2 } { x - 2 } > 0$