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Inequalities Involving Quotients Solve the nonlinear inequality. Express the solution using interval notation, and graph the solution set. $$\frac{x-4}{2 x+1} < 5$$

Short Answer

Expert verified
The solution is \((-\infty, -1) \cup (-\frac{1}{2}, \infty)\).

Step by step solution

01

Move all terms to one side of the inequality

Start by subtracting 5 from both sides of the inequality to get all terms on one side. This gives us: \( \frac{x-4}{2x+1} - 5 < 0 \).
02

Combine the fractions

Convert \(5\) to a fraction with the same denominator as \( \frac{x-4}{2x+1} \). Rewrite the inequality as: \( \frac{x-4 - 5(2x+1)}{2x+1} < 0 \).
03

Simplify the numerator

Simplify the numerator by distributing the \(-5\) and combining like terms: \[ (x-4) - (10x + 5) = x - 10x - 4 - 5 = -9x - 9 \]. This results in \( \frac{-9x - 9}{2x+1} < 0 \).
04

Determine the critical points

Identify the critical points where the numerator or denominator equals zero. Set \(-9x - 9 = 0\) and \(2x + 1 = 0\). Solving these gives the critical points \(x = -1\) and \(x = \frac{-1}{2}\).
05

Test intervals

The critical points divide the number line into intervals: \((-\infty, -1)\), \((-1, -\frac{1}{2})\), and \((-\frac{1}{2}, \infty)\). Pick a test point from each interval and substitute into \( \frac{-9x - 9}{2x+1} \) to determine whether the expression is positive or negative.
06

Analyze sign changes

Choose test points such as \(x = -2\) for \((-\infty, -1)\), \(x = -0.75\) for \((-1, -\frac{1}{2})\), and \(x = 0\) for \((-\frac{1}{2}, \infty)\). Substitute these values into the expression: - For \(x = -2\), \(\frac{-9(-2) - 9}{2(-2)+1} = \frac{18 - 9}{-4+1} = -3\), negative.- For \(x = -0.75\), \(\frac{-9(-0.75) - 9}{2(-0.75)+1} = \frac{6.75 - 9}{-1.5+1} = -\frac{9}{0.5}\), positive.- For \(x = 0\), \(\frac{-9(0) - 9}{2(0)+1} = \frac{-9}{1} = -9\), negative.
07

Identify the solution intervals

Since we want \(\frac{-9x - 9}{2x+1} < 0\), the intervals where the expression is negative are \((-\infty, -1)\) and \((-\frac{1}{2}, \infty)\). However, \(x = -\frac{1}{2}\) is not part of the solution because it makes the denominator zero. Thus, the solution is \((-\infty, -1) \cup (-\frac{1}{2}, \infty)\).
08

Express the solution in interval notation

Combine these intervals into interval notation. The final solution is \( (-\infty, -1) \cup (-\frac{1}{2}, \infty) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nonlinear Inequalities
When dealing with nonlinear inequalities, we are working with inequalities that contain terms of varying degrees. Unlike linear inequalities that involve straight-line expressions like \( ax + b \), nonlinear inequalities may have terms such as squares or fractions that make the inequality more complex. In other words, the relationship between the variables might not follow a straight line.

Solving nonlinear inequalities often involves manipulating the inequality to identify intervals on the number line where the inequality holds true. The use of algebraic techniques and understanding the behavior of the expression across different segments is key. We need to determine when the entire expression is positive or negative, which involves careful transformation and testing of specific values. Overall, dealing with nonlinear inequalities requires a combination of algebraic skills and conceptual insights.
Critical Points in Inequalities
Critical points play a crucial role when solving inequalities, particularly nonlinear ones. They are the values of the variable where the expression either equals zero or becomes undefined. These points help us divide the number line into intervals, each representing a potential range where the inequality might change its nature, from true to false or vice versa.

In this exercise, the critical points were identified by setting the numerator and denominator of the inequality to zero separately. For instance, from \(-9x - 9 = 0\) and \(2x + 1 = 0\), we found that the critical points are \(x = -1\) and \(x = -\frac{1}{2}\).
  • These points aren't part of our solution unless they make the inequality precisely zero or unless specifically included due to non-strict inequalities.
  • They help indicate where to test the expression in each interval.
Remember, identifying these points correctly is essential as they mark the transition areas for the inequality.
Interval Notation
Interval notation is a concise way of representing ranges of numbers that are solutions to inequalities. It helps easily communicate the solution set for an inequality without having to list out all potential solutions.
  • Parentheses \(( )\) denote that the endpoint is not included in the interval, used for strict inequalities like \(<\) or \(>\).
  • Brackets \([ ]\) are used when endpoints are included, in non-strict inequalities like \(\leq\) or \(\geq\).
In this solution, \((-\infty, -1) \cup (-\frac{1}{2}, \infty)\) indicates two separate ranges for \(x\) that satisfy the inequality. The use of the union symbol \( \cup \) means we are combining these ranges to represent the entire set of valid \(x\) values. Being fluent in interval notation is very useful, as it allows you to quickly read and understand the scope of solutions for an inequality.
Rational Expressions
Rational expressions involve fractions where the numerator and the denominator are polynomials. Because of the division, extra care is needed when handling these expressions, especially when solving inequalities.

In our exercise, the rational expression \(\frac{x-4}{2x+1}\) takes center stage. The inequality solution process includes:
  • Moving terms to create a zero inequality for easier manipulation.
  • Identifying valid ranges by avoiding division by zero, making sure to exclude points where the denominator is zero as these make the expression undefined.
  • Simplifying and solving the inequality requires ensuring that both the numerator and the denominator are considered in solving.
Understanding rational expressions is key to navigating complex inequalities. Successfully managing these expressions means consistently looking out for where operations may cause a denominator to equal zero, thereby dictating excluded values in the solution response.

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