Problem 31 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 32 , find the solution set of the given inequality and illustrate the solution on the real number $$ \frac{1}{3 x-7} \geq \frac{4}{3-2 x} $$

Step-by-Step Solution

Answer

The solution set is \( (-\infty, \frac{3}{2}) \cup (\frac{3}{2}, \frac{31}{14}] \).

Step 1: Identify the inequality

We need to solve the inequality \( \frac{1}{3x-7} \geq \frac{4}{3-2x} \).

Step 2: Find a common denominator

To compare the fractions, find a common denominator, which is \( (3x-7)(3-2x) \).

Step 3: Rewrite the inequality

Rewrite the inequality using the common denominator: \[ \frac{1(3-2x)}{(3x-7)(3-2x)} \geq \frac{4(3x-7)}{(3x-7)(3-2x)} \]

Step 4: Combine and simplify

Combine the terms on the left side and simplify: \[ \frac{3-2x}{(3x-7)(3-2x)} \geq \frac{12x-28}{(3x-7)(3-2x)} \]

Step 5: Clear the denominator

Since the denominator is the same, the inequality simplifies to: \[ 3-2x \geq 12x - 28 \]

Step 6: Solve for x

Solve the linear inequality: \[ 3 - 2x \geq 12x - 28 \] Move all terms involving x to one side and constants to the other: \[ 3 + 28 \geq 12x + 2x \] \[ 31 \geq 14x \] Divide by 14: \[ x \leq \frac{31}{14} \]

Step 7: Consider critical points

Check critical points from the original inequality. The denominators change signs when 3x-7=0 and 3-2x=0. These give critical points: \[ x = \frac{7}{3} \] and \[ x = \frac{3}{2} \]

Step 8: Test intervals

Test intervals formed by critical points in the inequality: \[ (-\infty, \frac{3}{2}), (\frac{3}{2}, \frac{7}{3}), (\frac{7}{3}, \infty) \]

Step 9: State the solution

After testing intervals, the solution set includes x values that do not make any denominator zero and satisfy \( x \leq \frac{31}{14} \). Therefore, the solution set is \[ (-\infty, \frac{3}{2}) \cup (\frac{3}{2}, \frac{31}{14}] \]

Key Concepts

inequalities

Inequalities are mathematical statements that compare two expressions. They use symbols such as greater than ( > ), less than ( < ), greater than or equal to ( \geq ), and less than or equal to ( \leq ). In the provided exercise, we worked with a rational inequality: \ \ \( \frac{1}{3x-7} \geq \frac{4}{3-2x} \).\ Inequality problems can be tricky because they involve finding the range of values for which the inequality holds true.\ \ When solving inequalities, follow these steps: \

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• Identify the inequality and rewrite it, if necessary.
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• Simplify both sides as much as possible.
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• Determine the critical points (places where the expression is undefined or equals zero).
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• Test the regions separated by the critical points.
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\ Always remember to flip the inequality sign when multiplying or dividing both sides by a negative number.

critical points

Critical points are key values that divide the real number line into separate regions. These points are found by setting the denominators or numerators of the fractions in the inequality to zero.\ \ In our exercise, the inequality \( \frac{1}{3x-7} \geq \frac{4}{3-2x} \) gives us critical points by solving:\ \

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• \( 3x - 7 = 0 \rightarrow x = \frac{7}{3} \)
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• \( 3 - 2x = 0 \rightarrow x = \frac{3}{2} \)
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\ At these points, the original rational expressions are undefined, and the inequality changes its behavior.\ \ Critical points help us determine where to test the intervals. By evaluating points within the intervals formed by critical points, we can verify which sections satisfy the inequality.

solution intervals

Solution intervals are parts of the number line where the inequality holds true based on critical points and interval testing. After determining the critical points \ \( x = \frac{7}{3} \) and \( x = \frac{3}{2} \), \ the real number line is split into three intervals: \ \

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• \( (-\infty, \frac{3}{2}) \)
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• \( (\frac{3}{2}, \frac{7}{3}) \)
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• \( (\frac{7}{3}, \infty) \)
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\ We test points in each interval within the inequality to see if they satisfy it. For instance: \ - Testing a point in \( (-\infty, \frac{3}{2}) \), say \( x = 0 \), shows this interval satisfies the inequality. \ - Testing a point in \( (\frac{3}{2}, \frac{7}{3}) \), say \( x = 2 \), shows that this interval does not satisfy the inequality. \ - Testing a point in \( (\frac{7}{3}, \infty) \), say \( x = 3 \), shows this interval does not satisfy the inequality either. \ \ Combining our results, our solution is: \ \( (-\infty, \frac{3}{2}) \cup (\frac{3}{2}, \frac{31}{14}] \). These intervals represent all \( x \) values where the inequality holds true.