Problem 55 - Algebra and Trigonometry

Question

\(33-66\) . 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 \((-\infty, 5) \cup [16, \infty)\).

Step 1: Bring the inequality to a single fraction

To solve the inequality \( \frac{2x+1}{x-5} \leq 3 \), start by subtracting 3 from both sides to bring everything under one fraction: \[ \frac{2x+1}{x-5} - 3 \leq 0 \]Rewriting 3 as a fraction with the same denominator gives:\[ \frac{2x+1}{x-5} - \frac{3(x-5)}{x-5} \leq 0 \]Combine the fractions:\[ \frac{2x+1 - 3(x-5)}{x-5} \leq 0 \]

Step 2: Simplify the numerator

Simplify the expression inside the numerator:\[ 2x + 1 - 3(x - 5) = 2x + 1 - 3x + 15 = -x + 16 \]So the inequality becomes:\[ \frac{-x + 16}{x-5} \leq 0 \]

Step 3: Identify the critical points

Set the numerator and the denominator equal to zero to find the critical points:\[-x + 16 = 0 \rightarrow x = 16 \]\[x - 5 = 0 \rightarrow x = 5 \]The critical points are \(x = 5\) and \(x = 16\).

Step 4: Test intervals

Divide the real number line into intervals using the critical points: - \((-\infty, 5)\)- \((5, 16)\)- \((16, \infty)\)Test each interval in the inequality \( \frac{-x + 16}{x-5} \leq 0 \):- For \( x = 0 \) in \((-\infty, 5)\), \( \frac{-0 + 16}{0 - 5} = -\frac{16}{5} < 0 \), so it works.- For \( x = 10 \) in \((5, 16)\), \( \frac{-10 + 16}{10 - 5} = \frac{6}{5} > 0 \), so it does not work.- For \( x = 20 \) in \((16, \infty)\), \( \frac{-20 + 16}{20 - 5} = -\frac{4}{15} < 0 \), so it works.

Step 5: Consider endpoint behavior

Consider the points \( x = 5 \) and \( x = 16 \):- At \( x = 5 \), the expression is undefined because of a zero in the denominator.- At \( x = 16 \), \( \frac{-16 + 16}{16 - 5} = 0 \), which satisfies the \( \leq 0 \) part.Include \( x = 16 \) but exclude \( x = 5 \).

Step 6: Write the solution in interval notation

Since the intervals \((-\infty, 5)\) and \((16, \infty)\) work, and including \(x = 16\), the solution is expressed in interval notation as:\[ (-\infty, 5) \cup [16, \infty) \]

Key Concepts

Critical Points

When solving a nonlinear inequality, identifying critical points is crucial. They are where the function changes behavior, such as sign changes or undefined points. For the inequality \( \frac{-x + 16}{x-5} \leq 0 \), we find critical points by setting both the numerator \(-x + 16\) and the denominator \(x-5\) equal to zero. This gives us:

• \(-x + 16 = 0 \rightarrow x = 16\)
• \(x - 5 = 0 \rightarrow x = 5\)

In this case, these points divide the number line into intervals. Understanding them helps in determining where the inequality may hold or not. Critical points can be potential solutions or boundaries for intervals where the inequality is defined.

Interval Notation

Interval notation is a simple way to express solutions of inequalities. It uses brackets and parentheses to show which parts of the number line satisfy the inequality. In the inequality \( \frac{-x + 16}{x-5} \leq 0 \), we end up with the solutions \((-\infty, 5)\) and \([16, \infty)\). Here’s how:

• The parenthesis \(( )\) indicate that an endpoint is not included in the solution set, like \(x = 5\) which is not part of the solution because the expression becomes undefined (zero in the denominator).
• The bracket \([ ]\) indicates that an endpoint is included; \(x = 16\) satisfies the \(\leq 0\) condition perfectly as it turns the expression to zero.

Interval notation provides a concise way to capture this information, making it easier to read and understand.

Inequality Solution

Solving a nonlinear inequality involves multiple steps and verifying intervals to see which portions of the number line satisfy the given inequality. For \( \frac{-x + 16}{x-5} \leq 0 \), we rearrange and simplify to find critical points, which reveal the intervals to test. Checking these intervals helps us understand where the inequality holds true:

• Come up with potential solutions by solving for critical points and determining significant intervals.
• Evaluate the inequality with test points from these intervals to see where the inequality is true.
• Analyze the solutions around critical points to include or exclude them in the final set.

Ultimately, the viable solutions in interval notation which solve this inequality are combined to give a comprehensive answer, indicating on which intervals the original inequality is true.

Test Intervals

To ensure a reliable solution, you must test intervals derived from critical points. This involves selecting test values from each interval and plugging them back into the inequality:\( \frac{-x + 16}{x-5} \leq 0 \). Here's a breakdown:

• For the interval \((-\infty, 5)\), test with a value like \(x = 0\), resulting in \(-\frac{16}{5} < 0\) which satisfies the inequality.
• Use a midpoint in \((5, 16)\), like \(x = 10\), which gives \(\frac{6}{5} > 0\), showing it does not satisfy the inequality.
• Try \(x = 20\) for \((16, \infty)\), yielding \(-\frac{4}{15} < 0\), meeting the inequality criteria.

Testing these intervals helps confirm which sections of the number line provide valid solutions. This step is crucial for accurately determining where the inequality holds.