Problem 53 - Intermediate Algebra

Question

Solve each inequality. $$ \frac{3 x+2}{x+4} \leq 2 $$

Step-by-Step Solution

Answer

The solution is \( x \in (-4, 6] \).

Step 1: Move Everything to One Side

Subtract 2 from both sides to set the inequality to zero:\[ \frac{3x+2}{x+4} - 2 \leq 0 \]This will help us find the critical points where the inequality might equal zero.

Step 2: Bring to Common Denominator

Rewrite 2 with the same denominator:\[ \frac{3x+2}{x+4} - \frac{2(x+4)}{x+4} \leq 0 \]This merges the terms under a single fraction.

Step 3: Simplify the Numerator

Simplify the expression in the numerator:\[ \frac{3x+2 - 2x - 8}{x+4} \leq 0 \]Combine the numerators:\[ \frac{x - 6}{x+4} \leq 0 \]

Step 4: Identify Critical Points

Solve for the values when the numerator and denominator are zero:- Numerator zero: \( x - 6 = 0 \) \( \Rightarrow x = 6 \)- Denominator zero: \( x + 4 = 0 \) \( \Rightarrow x = -4 \)These points divide the real number line into intervals to test the inequality.

Step 5: Test Intervals

The critical points divide the number line into three intervals: \((-\infty, -4)\), \((-4, 6)\), and \((6, \,\infty)\).- Choose a test point in each interval and substitute it into \( \frac{x - 6}{x+4} \).- For \((-\infty, -4)\), choose \( x = -5 \): \( \frac{-5 - 6}{-5 + 4} = \frac{-11}{-1} = 11 > 0 \)- For \((-4, 6)\), choose \( x = 0 \): \( \frac{0 - 6}{0 + 4} = \frac{-6}{4} = -1.5 < 0 \)- For \((6, \,\infty)\), choose \( x = 7 \): \( \frac{7 - 6}{7 + 4} = \frac{1}{11} > 0 \).This indicates the inequality is satisfied in the interval \((-4, 6)\).

Step 6: Determine Inclusion of Critical Points

Check which critical points satisfy the inequality \( \frac{x - 6}{x+4} = 0 \) or are undefined.- \( x = 6 \) makes the fraction zero, included in \([-4, 6] \).- \( x = -4 \) makes the denominator zero and is undefined, not included.Thus, the solution includes the endpoint 6 but not \(-4\).

Key Concepts

Critical Points

When solving inequalities like \( \frac{3x+2}{x+4} \leq 2 \), identifying the critical points is crucial. Critical points occur when the numerator or denominator equals zero. These determine where the expression changes sign.
To find the critical points:

• Set the numerator to zero: \( x - 6 = 0 \Rightarrow x = 6 \).
• Set the denominator to zero: \( x + 4 = 0 \Rightarrow x = -4 \).

Critical points here are \( x = -4 \) and \( x = 6 \). These divide the number line into intervals. Understanding these points will guide which regions satisfy the inequality.

Numerical Intervals

Once you identify the critical points, they divide the number line into numerical intervals. These intervals are segments between and beyond these points. For our example, the critical points \( x = -4 \) and \( x = 6 \) create three intervals:

• \((-\infty, -4)\)
• \((-4, 6)\)
• \((6, \infty)\)

Each interval needs to be tested for inequality satisfaction. Numerical intervals show where conditions of the inequality are valid, helping isolate the solution set.

Numerator Simplification

In order to solve the inequality \( \frac{3x+2}{x+4} \leq 2 \), it's necessary to simplify the numerator after rearranging terms. Start by expressing the inequality with a single fraction:

• Subtract 2 expressed as a fraction: \( \frac{2(x+4)}{x+4} \).
• This becomes \( \frac{3x+2 - 2x - 8}{x+4} \).
• Simplify the arithmetic: \( \frac{x - 6}{x+4} \).

Simplifying the numerator helps locate where the inequality holds true by pinpointing the zero or undefined aspects of fractions.

Test Intervals

Testing intervals helps verify which segments fulfill the inequality. After dividing the number line, pick a test value from each interval:

• For \((-\infty, -4)\), use \( x = -5 \): \( \frac{-11}{-1} = 11 > 0 \).
• For \((-4, 6)\), use \( x = 0 \): \( \frac{-6}{4} = -1.5 < 0 \).
• For \((6, \infty)\), use \( x = 7 \): \( \frac{1}{11} > 0 \).

These tests determine which intervals satisfy \( \frac{x - 6}{x+4} \leq 0 \). Here, only \((-4, 6)\) is valid, showing that critical assessment of intervals refines the solution.