Question

Solve each inequality algebraically. $$\frac{3 x-5}{x+2} \leq 2$$

Short answer

The solution is \(-2 < x \leq 9\).

Step-by-step solution

Move the constant to the other side

Subtract 2 from both sides of the inequality: \[ \frac{3x - 5}{x + 2} - 2 \leq 0 \]

Find a common denominator

Combine the terms on the left by finding a common denominator (which is \(x + 2\)): \[ \frac{3x - 5 - 2(x + 2)}{x + 2} \leq 0 \]

Simplify the numerator

Distribute and simplify the terms in the numerator: \[ \frac{3x - 5 - 2x - 4}{x + 2} \leq 0 \frac{x - 9}{x + 2} \leq 0 \]

Determine the critical points

Set the numerator and denominator equal to zero to find the critical points: \[ x - 9 = 0 \implies x = 9 x + 2 = 0 \implies x = -2 \]

Test the intervals

Test the intervals around the critical points \(-\infty, -2\), \(-2, 9\), and \(9, \infty\). Choose test points in each interval and substitute into the inequality to determine where the inequality holds.

Write the solution set

Based on the interval testing: The inequality holds for \( -2 < x \leq 9 \).

Key concepts

Inequality solving

Solving inequalities can initially be tricky, but once you understand the steps, it becomes much easier. Inequalities are similar to equations, but instead of an equal sign, they use inequality symbols like \( \leq \), \( \geq \), \( < \), and \( > \). Our goal is to find the range of values that satisfy the inequality.
Let's start by moving all terms to one side of the inequality so that one side is zero. This often means subtracting or adding terms to both sides.
For the example \( \frac{3 x - 5}{x + 2} \leq 2 \), we subtract 2 from both sides to get: \[ \frac{3 x - 5}{x + 2} - 2 \leq 0 \]
By isolating the terms on one side, we can interpret the inequality correctly and prepare for further simplification.

Critical points

To solve an inequality, identifying the critical points is a crucial step. Critical points divide the number line into segments where the inequality may change from true to false or vice versa.
Critical points are values of \(x\) that make either the numerator or the denominator zero. For the given inequality \( \frac{x - 9}{x + 2} \leq 0 \), we set each part to zero:

• For the numerator: \( x - 9 = 0 \implies x = 9 \)
• For the denominator: \( x + 2 = 0 \implies x = -2 \)

These are our critical points. We use them to determine the intervals to test where the inequality holds. Knowing your critical points helps you understand where any inequality expressions might change behavior.

Common denominators

Finding a common denominator is an important technique when working with fractions in inequalities. A common denominator allows us to combine terms effectively.
In the problem, after subtracting 2 from both sides, we have: \( \frac{3x - 5 - 2(x + 2)}{x + 2} \leq 0 \). To simplify this, distribute the \( 2 \) and combine like terms:

• \( 3x - 5 - 2x - 4 = x - 9 \)

This gives us: \[ \frac{x - 9}{x + 2} \leq 0 \]
Combining terms under a common denominator enables us to work with a single fraction and single inequality, making the subsequent steps more straightforward.

Interval testing

Interval testing is the final step in solving inequality problems. It helps determine where the inequality holds true. Once we have our critical points, we break the number line into intervals.
For the critical points \( -2 \) and \( 9 \), the intervals to test are:

• \text{-∞, -2}
• \text{-2, 9}
• \text{9, ∞}

Choose a test point in each interval:

• \text{For}\text{-3}\text{in}\text{\text{-∞, -2}}: \( \frac{-3 - 9}{-3 + 2} = \frac{-12}{-1} = 12 \rightarrow \text{not} \leq 0 \)
• \text{For}\text{0}\text{in}\text{\text{-2, 9}}: \( \frac{0 - 9}{0 + 2} = \frac{-9}{2} = -4.5 \rightarrow \leq 0 \)
• \text{For}\text{10}\text{in}\text{\text{9, ∞}}: \( \frac{10 - 9}{10 + 2} = \frac{1}{12} \rightarrow \text{not} \leq 0 \)

Based on this testing, the inequality holds true in \text{-2, 9}. Make sure to also consider endpoint behavior. The final solution is \(-2 < x \leq 9 \).