Problem 28 - Algebra A Combined Function

Question

Solve each inequality. Write the solution set in interval notation. $$ (6 x+7)(7 x-12)>0 $$

Step-by-Step Solution

Answer

The solution set is \((-\infty, -\frac{7}{6}) \cup (\frac{12}{7}, \infty)\).

Step 1: Expand the Inequality

First, we need to examine the inequality \((6x + 7)(7x - 12) > 0\). Notice that it is in a factored form, so we don't need to expand it further. This inequality is already set to find intervals where the product is positive.

Step 2: Find the Critical Points

Set each factor equal to zero to find the critical points: \(6x + 7 = 0\) and \(7x - 12 = 0\). Solving these gives:- For \(6x + 7 = 0\), we have \(x = -\frac{7}{6}\).- For \(7x - 12 = 0\), we have \(x = \frac{12}{7}\).

Step 3: Analyze the Intervals

Use the critical points to determine the intervals we're analyzing. These points divide the number line into three intervals:1. \((-\infty, -\frac{7}{6})\)2. \((-\frac{7}{6}, \frac{12}{7})\)3. \((\frac{12}{7}, \infty)\).

Step 4: Test Each Interval

In each interval, choose a test value and determine if it makes the inequality true:- For \((-\infty, -\frac{7}{6})\), test \(x = -2\): \((6(-2) + 7)(7(-2) - 12) = (-5)(-26) = 130 > 0\)- For \((-\frac{7}{6}, \frac{12}{7})\), test \(x = 0\): \((6(0) + 7)(7(0) - 12) = (7)(-12) = -84 < 0\)- For \((\frac{12}{7}, \infty)\), test \(x = 2\): \((6(2) + 7)(7(2) - 12) = (19)(2) = 38 > 0\)

Step 5: Write the Solution in Interval Notation

Since the inequality is true for \((-\infty, -\frac{7}{6})\) and \((\frac{12}{7}, \infty)\), the solution set in interval notation is:\((-\infty, -\frac{7}{6}) \cup (\frac{12}{7}, \infty)\).

Key Concepts

Factoring

Factoring is a fundamental skill in algebra that helps simplify expressions and solve equations or inequalities. When dealing with inequalities, like the one we have here \((6x + 7)(7x - 12) > 0\), factoring means breaking down the expression into the product of simpler terms or factors. In this particular exercise, we are fortunate as the inequality is already in factored form. Factoring reveals important roots or critical points of the inequality which serve as boundaries for intervals. Once the inequality is factored, we can more easily identify the values of \(x\) that will make the inequality true or false.
\(-\frac{7}{6}\) and \(\frac{12}{7}\) are the roots, which we find by solving \(6x + 7 = 0\) and \(7x - 12 = 0\). These roots are crucial in determining the intervals we'll later work with. Overall, factoring is like setting up the stage for systematically examining the inequality's behavior.

Interval Notation

Interval notation is essential for expressing the solution set of inequalities. It provides a clear and concise way to describe sets of numbers that represent the solution. In our example, after determining where our inequality holds true, we use interval notation to express this clearly.
The critical points \(-\frac{7}{6}\) and \(\frac{12}{7}\) help us form intervals

• \((-\infty, -\frac{7}{6})\)
• \((\frac{12}{7}, \infty)\)

These intervals represent sections of the number line where the inequality \((6x + 7)(7x - 12) > 0\) is satisfied. In interval notation, parentheses \(()\) indicate that endpoints are not included in the interval, which is important in showing that the inequality does not hold true at the critical points themselves. Thus, capturing the solution set in interval notation is a streamlined process that showcases where the inequality remains positive.

Test Interval Method

The test interval method is a logical technique for determining where an inequality holds true along various sections of the number line. After factoring, we use the critical points to identify intervals, such as \((-\infty, -\frac{7}{6})\), \((-\frac{7}{6}, \frac{12}{7})\), and \((\frac{12}{7}, \infty)\).
Our goal is to test each interval with a representative value to see if it satisfies the inequality. For example:

• In \((-\infty, -\frac{7}{6})\), we test \(x = -2\): \((6(-2) + 7)(7(-2) - 12) > 0\), which confirms positivity.
• In \((-\frac{7}{6}, \frac{12}{7})\), testing \(x = 0\) results in a negative value, hence it does not satisfy the inequality.
• In \((\frac{12}{7}, \infty)\), checking \(x = 2\) results in a positive value, confirming it meets the inequality condition.

By applying a test value in each defined interval, this method gives us a clear picture of where the inequality is valid and helps craft a precise solution using interval notation.