Math Problem Statement

-2x+1≤0 over x-3

Solution

The inequality you provided is:

2x+1x30\frac{-2x + 1}{x - 3} \leq 0

To solve this inequality, let's break it down into a step-by-step process.

Step 1: Identify the critical points

We need to find the values of xx that make the expression either 0 or undefined:

  1. Set the numerator equal to 0: 2x+1=0    x=12-2x + 1 = 0 \implies x = \frac{1}{2}

  2. Set the denominator equal to 0 (where the expression is undefined): x3=0    x=3x - 3 = 0 \implies x = 3

So, the critical points are x=12x = \frac{1}{2} and x=3x = 3.

Step 2: Analyze the sign of the expression

Now, we need to determine the sign of the expression on the intervals created by these critical points: (,12)(-\infty, \frac{1}{2}), (12,3)(\frac{1}{2}, 3), and (3,)(3, \infty).

  • For x(,12)x \in (-\infty, \frac{1}{2}), pick a test point like x=0x = 0: 2(0)+103=13=13\frac{-2(0) + 1}{0 - 3} = \frac{1}{-3} = -\frac{1}{3} This is negative.

  • For x(12,3)x \in (\frac{1}{2}, 3), pick a test point like x=1x = 1: 2(1)+113=2+12=12=12\frac{-2(1) + 1}{1 - 3} = \frac{-2 + 1}{-2} = \frac{-1}{-2} = \frac{1}{2} This is positive.

  • For x(3,)x \in (3, \infty), pick a test point like x=4x = 4: 2(4)+143=8+11=7\frac{-2(4) + 1}{4 - 3} = \frac{-8 + 1}{1} = -7 This is negative.

Step 3: Check the inequality

We want the expression to be less than or equal to 0, so we are looking for intervals where the expression is negative or zero.

  • The expression is negative on the intervals (,12)(-\infty, \frac{1}{2}) and (3,)(3, \infty).
  • The expression is zero at x=12x = \frac{1}{2}.

Step 4: Conclusion

The solution to the inequality is: x(,12](3,)x \in (-\infty, \frac{1}{2}] \cup (3, \infty)

This is the set of values for xx that satisfy the inequality.

Would you like further details or have any questions?

Here are 5 related questions to expand on this:

  1. How do you solve rational inequalities involving different expressions?
  2. What happens when both the numerator and denominator have higher-degree terms?
  3. How do vertical asymptotes affect the solution to inequalities?
  4. Can inequalities have multiple valid solution intervals?
  5. What are the properties of rational expressions when analyzing their sign?

Tip: Always carefully handle points where the expression is undefined, as these create important boundaries in rational inequalities.

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Math Problem Analysis

Mathematical Concepts

Rational Inequalities
Critical Points
Sign Analysis

Formulas

Solving Rational Inequalities
Critical Point Identification

Theorems

Rational Inequality Theorem
Sign Change Method

Suitable Grade Level

Grades 10-12