Problem 50 - College Algebra

Question

Solve each rational inequality in Exercises \(43-60\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{3 x+5}{6-2 x} \geq 0 $$

Step-by-Step Solution

Answer

The solution of \(\frac{3x+5}{6-2x} ≥ 0\) is expressed in interval notation as \((-∞,-\frac{5}{3}) ∪ (3, ∞)\).

Step 1: Bring the inequality to standard form

The inequality is already in standard form \(\frac{P(x)}{Q(x)} ≥ 0\), where \(P(x) = 3x+5\) and \(Q(x) = 6-2x\). If it was not, it would be necessary to manipulate the inequality so that one side is 0.

Step 2: Find critical values

Next, identify the values of x that make \(P(x) = 0\) or \(Q(x) = 0\). These are called the critical values. For \(P(x) = 3x+5\), if we solve for \(x\), we get \(x = \frac{-5}{3}\). For \(Q(x) = 6-2x\), \(x = 3\). So, the critical values are \(x = \frac{-5}{3}\) and \(3\). These values divide the real number line into several intervals.

Step 3: Analyze the intervals

Now examine the intervals determined by the critical values. Pick a test point in each interval and substitute that into the inequality. The intervals to test are \(-∞ < x < \frac{-5}{3}\), \(\frac{-5}{3} < x < 3\), and \(3 < x < ∞\). If the inequality is true for the test point, then it is true for all numbers in that interval. After the test, we find that the solution is \(-∞ < x < \frac{-5}{3}\) and \(3 < x < ∞\).

Step 4: Express the solution set in interval notation and graph the solution set on a real number line

Finally, we express the solution set as \((-∞,\frac{-5}{3}) ∪ (3, ∞)\) in interval notation. For the graphical representation on a real number line, we plot two open intervals, the first from negative infinity to \frac{-5}{3} (not including \rac{-5}{3}), and the second from 3 to positive infinity (not including 3).

Key Concepts

Interval Notation

Interval notation is a mathematical shorthand used to describe sets of numbers, especially those that are solutions to inequalities. It is particularly useful for expressing solution sets on the real number line without listing every single element. The format consists of two end-points and an indication of whether the endpoints are included or not.
Here are the basic symbols used:

• Parentheses, \((\) and \(),\) indicate that endpoints are not included in the set, meaning the set is open at that end.
• Brackets, \([\) and \()]\) indicate that endpoints are included, which makes the set closed at that end.
• Infinity, represented by \(\infty\), signifies an unbounded limit in a positive or negative direction and is always accompanied by a parenthesis, as infinity cannot be "included" in an interval.

In the provided exercise, the solution set expressed in interval notation is \((-\infty, \frac{-5}{3}) \cup (3, \infty)\). This means that all real numbers less than \(\frac{-5}{3}\) and greater than 3 are part of the solution, but \(\frac{-5}{3}\) and 3 themselves are not included. This is expressed by the use of parentheses around these endpoints.

Critical Values

Critical values are the points where the inequality is potentially equal to zero or undefined. Finding these values is a crucial step when solving rational inequalities because they divide the real number line into distinct intervals. Analyzing these intervals helps determine where the inequality holds true.
To find critical values:

• Set the numerator of the rational expression equal to zero and solve for \(x\). This gives points where the expression might change sign.
• Set the denominator equal to zero and solve for \(x\). This helps find points where the expression is undefined.

In the inequality \(\frac{3x + 5}{6 - 2x} \geq 0\), the numerator \(3x + 5 = 0\) gives a critical value of \(x = \frac{-5}{3}\), and the denominator \(6 - 2x = 0\) gives \(x = 3\). Dividing the real number line into intervals using these critical values allows us to test each interval and identify where the original inequality holds true.

Real Number Line

The real number line is a graphical representation of all possible real numbers, stretching infinitely in both negative and positive directions. When solving inequalities, the real number line is used to visually demonstrate the solution by marking intervals where the inequality is satisfied.
To graph a solution set on the real number line:

• Identify and mark the critical values on the line. These points are found by solving the numerator and denominator for zero in rational inequalities.
• Determine which intervals make the inequality true, often by picking test points from each interval.
• Draw an open circle at critical values if they are not included in the solution, and a closed circle if they are included.

For the exercise \(\frac{3x + 5}{6 - 2x} \geq 0\), the solution set divided by the critical points includes \((-\infty, \frac{-5}{3})\) for numbers less than \(\frac{-5}{3}\) and \((3, \infty)\) for numbers greater than 3. These can be represented on the real number line using open circles at \(\frac{-5}{3}\) and 3, highlighting the excluded endpoints.