Solve the rational inequality and graph the solution set on a real number line. Express the solution set in interval notation. [tex]\frac{8 x-3}{5 x-3} \leq 2[/tex] Solve the inequality. What is the solution set? Select the correct choice below and, if necessary, fill in the answer box. A. The solution set is [ ]. (Simplify your answer. Type your answer in interval notation. Type an exact answer, using radicals as needed. Use any numbers in the expression.) B. The solution set is the empty set.
Answer (2)
The solution to the inequality 5 x − 3 8 x − 3 ≤ 2 is ( − ∞ , 5 3 ) ∪ [ 2 3 , ∞ ) . This includes values where the expression is non-negative, excluding the point where the denominator is zero. Therefore, the correct choice is A.
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Subtract 2 from both sides and combine the fraction: 5 x − 3 8 x − 3 − 2 ≤ 0 ⇒ 5 x − 3 − 2 x + 3 ≤ 0 .
Multiply by -1 and flip the inequality sign: 5 x − 3 2 x − 3 ≥ 0 .
Find critical values by setting the numerator and denominator to zero: x = 2 3 and x = 5 3 .
Test intervals and write the solution in interval notation: ( − ∞ , 5 3 ) ∪ [ 2 3 , ∞ ) .
Explanation
Problem Analysis We are given the rational inequality 5 x − 3 8 x − 3 ≤ 2 . Our goal is to solve for x and express the solution set in interval notation.
Subtracting 2 First, we subtract 2 from both sides of the inequality to get 5 x − 3 8 x − 3 − 2 ≤ 0 .
Combining Terms Next, we find a common denominator and combine the terms: 5 x − 3 8 x − 3 − 2 ( 5 x − 3 ) ≤ 0 5 x − 3 8 x − 3 − 10 x + 6 ≤ 0 5 x − 3 − 2 x + 3 ≤ 0
Multiplying by -1 To make it easier to work with, we multiply both sides by -1. Remember to flip the inequality sign when multiplying by a negative number: 5 x − 3 2 x − 3 ≥ 0
Finding Critical Values Now, we find the critical values by setting the numerator and denominator equal to zero: 2 x − 3 = 0 ⇒ x = 2 3 5 x − 3 = 0 ⇒ x = 5 3
Creating Intervals We use these critical values to create intervals on the number line: ( − ∞ , 5 3 ) , ( 5 3 , 2 3 ) , and ( 2 3 , ∞ ) . We need to test a value in each interval to determine the sign of the expression 5 x − 3 2 x − 3 .
Testing Intervals Let's test the intervals:
For ( − ∞ , 5 3 ) , let x = 0 . Then 0"> 5 ( 0 ) − 3 2 ( 0 ) − 3 = − 3 − 3 = 1 > 0 . So, this interval is part of the solution.
For ( 5 3 , 2 3 ) , let x = 1 . Then 5 ( 1 ) − 3 2 ( 1 ) − 3 = 2 − 1 < 0 . So, this interval is not part of the solution.
For ( 2 3 , ∞ ) , let x = 2 . Then 0"> 5 ( 2 ) − 3 2 ( 2 ) − 3 = 7 1 > 0 . So, this interval is part of the solution.
Writing the Solution Set Since the inequality is ≥ 0 , we include the value where the numerator is zero, which is x = 2 3 . However, we must exclude the value where the denominator is zero, which is x = 5 3 , because the expression is undefined there. Therefore, the solution set is ( − ∞ , 5 3 ) ∪ [ 2 3 , ∞ ) .
Final Answer The solution set is ( − ∞ , 5 3 ) ∪ [ 2 3 , ∞ ) .
Examples
Rational inequalities are useful in various real-world scenarios. For instance, they can be used to model the concentration of a drug in the bloodstream over time. Suppose the concentration must remain below a certain level to avoid toxic effects. Solving a rational inequality can help determine the time intervals during which the drug concentration is safe and effective. Another example is in business, where rational inequalities can help determine the production levels needed to maintain a certain profit margin, considering factors like cost and revenue. Understanding and solving rational inequalities provides valuable tools for making informed decisions in these and other practical situations.
