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Magazine Chapter 8: Problem 28
Solve each inequality. Write the solution set in interval notation. $$(6 x+7)(7 x-12)>0$$
Short Answer
Expert verified
Solution: \((-\infty, -\frac{7}{6}) \cup (\frac{12}{7}, \infty)\).
Step by step solution
01
Find the Critical Values
First, we need to determine the critical values for the inequality \( (6x+7)(7x-12)>0 \). Set each factor equal to zero: \( 6x + 7 = 0 \) and \( 7x - 12 = 0 \). Solving these equations gives the critical values.\[ 6x + 7 = 0 \Rightarrow x = -\frac{7}{6} \] \[ 7x - 12 = 0 \Rightarrow x = \frac{12}{7} \]
02
Determine Intervals Using Critical Values
The critical values \( x = -\frac{7}{6} \) and \( x = \frac{12}{7} \) divide the number line into three intervals: \( (-\infty, -\frac{7}{6}) \), \( (-\frac{7}{6}, \frac{12}{7}) \), and \( (\frac{12}{7}, \infty) \).
03
Test Each Interval
Pick test points from each interval to determine where the expression is positive. 1. For \( (-\infty, -\frac{7}{6}) \), choose \( x = -2 \):\( (6(-2) + 7)(7(-2) - 12) = (-5)(-26) > 0 \). The inequality is true. 2. For \( (-\frac{7}{6}, \frac{12}{7}) \), choose \( x = 0 \):\( (6(0) + 7)(7(0) - 12) = 7(-12) < 0 \). The inequality is false. 3. For \( (\frac{12}{7}, \infty) \), choose \( x = 2 \):\( (6(2) + 7)(7(2) - 12) = (19)(2) > 0 \). The inequality is true.
04
Write the Solution in Interval Notation
The solution sets for the intervals where the inequality holds true are \((-finity, -\frac{7}{6})\) and \((\frac{12}{7}, \infty)\). Therefore, the solution in interval notation is \((-\infty, -\frac{7}{6}) \cup (\frac{12}{7}, \infty)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Values
Critical values are fundamental in solving rational inequalities. These are the x-values where the expression changes sign, typically found by setting each factor in an inequality to zero. For the expression \((6x + 7)(7x - 12) > 0\), you first set \(6x + 7 = 0\) and \(7x - 12 = 0\). Solving for x gives you the critical values, \(x = -\frac{7}{6}\) and \(x = \frac{12}{7}\).
- The critical values split the number line into intervals that need to be tested for the inequality.
- This process helps in identifying where the original inequality holds true or false.
Interval Notation
Interval notation is used to express a range of values compactly and is essential for representing solutions of inequalities. It involves using parentheses \((\) and brackets \([\) to denote the start and end of an interval on the number line. For our case, after finding critical values \((-\frac{7}{6}, \frac{12}{7})\), we determined the solution sets of the inequality lie within distinct intervals.
- Use parenthesis \((\) to indicate values that aren't included in the interval, while brackets \([\) include the endpoint.
- For the inequality \((6x + 7)(7x - 12) > 0\), the solution is \(( -\infty, -\frac{7}{6}) \cup (\frac{12}{7}, \infty)\).
Test Points
Choosing test points within intervals helps in verifying whether the intervals make an inequality true or false. When assessing \((6x + 7)(7x - 12) > 0\), after determining intervals using critical values, you choose simple numbers from each interval.
- For \((-\infty, -\frac{7}{6})\), you might pick \(x = -2\) to test if this makes the inequality hold true.
- For \((-\frac{7}{6}, \frac{12}{7})\), choosing \(x = 0\) provides a simple calculation.
- Finally, testing \(x = 2\) in \((\frac{12}{7}, \infty)\) confirms if this stretch of the number line solves the inequality.
Rational Inequalities
Rational inequalities involve ratios of polynomials and understanding them is key to problems like \((6x + 7)(7x - 12) > 0\). They require breaking down the expression into parts, where each polynomial is set against zero to find critical points and then analyzing intervals.
- Such inequalities may involve sign changes between intervals divided by zero points, identified as critical values.
- Test points then allow checking which regions satisfy the inequality condition, positive or negative.
- It goes beyond numerical solutions by fostering deeper analytical skills and understanding of polynomial behavior.
