Problem 61 - Algebra and Trigonometry

Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ \frac{2 x+1}{x-5} \leq 3 $$

Step-by-Step Solution

Answer

The solution is \([16, \infty)\).

Step 1: Bring Inequality to Standard Form

First, we need to bring the inequality to a standard form by subtracting 3 from both sides:\[\frac{2x+1}{x-5} - 3 \leq 0\]To combine the terms, convert 3 into a fraction with the same denominator by rewriting it as \(\frac{3(x-5)}{x-5}\), resulting in the inequality:\[\frac{2x+1 - 3(x-5)}{x-5} \leq 0\]

Step 2: Simplify the Expression

Simplify the numerator of the fraction:\[2x + 1 - 3x + 15 = -x + 16\]So, the inequality becomes:\[\frac{-x + 16}{x - 5} \leq 0\]Rewriting it:\[\frac{-(x - 16)}{x - 5} \leq 0\]

Step 3: Find Critical Points

The function \(f(x) = \frac{-(x - 16)}{x - 5}\) is undefined at \(x = 5\) and changes sign at the zeros of the numerator, \(x = 16\). These points, \(x = 5\) and \(x = 16\), are the critical points that will divide the number line into intervals.

Step 4: Test Intervals on the Number Line

Identify the intervals determined by the critical points: \((-\infty, 5)\), \((5, 16)\), and \((16, \infty)\). Choose a test point from each interval to determine the sign of \(f(x)\):- For \((-\infty, 5)\), choose \(x = 0\): \(f(0) = \frac{-(0 - 16)}{0 - 5} = \frac{-16}{-5} > 0\).- For \((5, 16)\), choose \(x = 10\): \(f(10) = \frac{-(10 - 16)}{10 - 5} = \frac{-(-6)}{5} > 0\).- For \((16, \infty)\), choose \(x = 20\): \(f(20) = \frac{-(20 - 16)}{20 - 5} = \frac{-4}{15} < 0\).

Step 5: Determine the Solution Set

The inequality \(\frac{-(x-16)}{x-5} \leq 0\) holds when \(f(x)\leq 0\). From testing, it is true on the interval \((16, \infty)\). Notice that at \(x = 16\), \(f(x) = 0\), so \(x = 16\) is part of the solution, giving us \([16, \infty)\). The point \(x = 5\) is excluded because it makes the denominator zero.

Step 6: Graph the Solution Set

Graph the interval \([16, \infty)\) on the number line, indicating that the solution includes the point \(x = 16\) with a closed dot and extends indefinitely to the right. No solutions exist for \(x = 5\) to \(x = 16\) and from \(-\infty\) to \(5\), which remain unshaded.

Key Concepts

Interval Notation

Interval notation is a way of representing subsets of real numbers, particularly suited for describing solutions to inequalities. It efficiently conveys the start and end points of an interval, and indicates whether those end points are included or excluded from the interval.

In interval notation, brackets and parentheses do the heavy lifting:

• Square brackets \( [ \text{ or } ] \) indicate that an endpoint is included in the interval. This is also known as a "closed" interval.
• Parentheses \( ( \text{ or } ) \) show that an endpoint is not included. This is referred to as an "open" interval.

Let's consider the interval \[ [16, \infty) \] from the solution set of the inequality problem. Here, the square bracket next to 16 indicates that 16 is included in the solution (because at this point, the function equals zero, satisfying the inequality). On the other hand, infinity (\( \infty \)), can never be included, as it's a concept rather than a specific number, hence the parenthesis.

Proper understanding of interval notation is key to correctly expressing solutions to inequalities.

Numerators and Denominators

In mathematics, fractions play an essential role in expressing parts of a whole. For inequalities involving fractions, such as \( \frac{-(x - 16)}{x - 5} \leq 0 \), understanding numerators and denominators is crucial.

The numerator is the top part of a fraction. Here, it is \(-(x - 16)\). This part of the expression determines when the entire fraction is zero or changes signs. In our exercise, the numerator zeroes out when \( x = 16 \), making it a critical point.

The denominator, \( x - 5 \), is the bottom part of a fraction. It's important to remember that a fraction becomes undefined if its denominator is zero. When \( x = 5 \), the denominator becomes zero, meaning the inequality can't be evaluated at this point.

By simplifying the numerator and ensuring the denominator doesn't equal zero, we efficiently solve and interpret inequalities involving fractions.

Critical Points

Critical points in solving inequalities are where the function can change its behavior or isn't defined. They mark boundaries on the number line to test distinct intervals. To identify critical points, focus on:

• Where the numerator equals zero.
• Where the denominator is undefined.

For the function \( f(x) = \frac{-(x - 16)}{x - 5} \), the critical points are determined as follows:
1. Set the numerator equal to zero: \(-(x - 16) = 0\). This simplifies to \(x = 16\). At this point, the function value is zero.
2. Find where the denominator is zero: \(x - 5 = 0\). This gives \(x = 5\), where the function is undefined.
These critical points are fundamental in breaking down the number line into intervals like \((\infty, 5)\), \((5, 16)\), and \((16, \infty)\). Testing sign changes within these intervals allows us to determine where the inequality holds true. Knowing this, we accurately derive the solution set, ensuring valid intervals are identified.