Problem 86 - Precalculus Student Solutions Manual 5th

Question

Solve each rational inequality. Write each solution set in interval notation. \(4\frac{x+2}{3+2 x} \leq 5\)4

Step-by-Step Solution

Answer

The solution set is: \( \left[ \frac{-7}{6}, +\infty \right) \setminus \{-1.5\} \).

Step 1: Simplify the Inequality

Start by rewriting the given inequality: \(4 \frac{x+2}{3+2x} \leq 5\). Distribute the 4 to get: \( \frac{4(x+2)}{3+2x} \leq 5\).

Step 2: Eliminate the Fraction

Multiply both sides by \(3 + 2x\) to eliminate the denominator: \(4(x + 2) \leq 5(3 + 2x)\) Be mindful that the inequality will flip if \(3 + 2x\) is negative.

Step 3: Distribute and Combine Like Terms

Distribute on both sides: \(4x + 8 \leq 15 + 10x\). Subtract \(4x\) from both sides: \(8 \leq 15 + 6x\).

Step 4: Solve for x

Subtract 15 from both sides to get: \(8 - 15 \leq 6x\), which simplifies to: \(-7 \leq 6x\). Finally, divide both sides by 6: \(\frac{-7}{6} \leq x\Rightarrow x \geq \frac{-7}{6}\).

Step 5: Consider the Denominator's Restriction

Recall \(3 + 2x eq 0\), so solve \(3 + 2x = 0\): \(2x = -3 \ x = -1.5\).

Step 6: Combine Both Conditions

Since \(x \geq \frac{-7}{6}\) and \(x eq -1.5\), the solution set is: \[ \left[ \frac{-7}{6}, +\infty \right) \setminus \{-1.5\} \].

Key Concepts

Rational Inequalities

Rational inequalities involve expressions containing fractions with polynomials in the numerator and the denominator. These expressions are compared using inequality symbols such as <, >, ≤, and ≥. Solving them requires careful manipulation of both the numerator and denominator.
To solve rational inequalities, ensure the inequality is in a format that isolates the rational expression on one side and 0 on the other.
For example, an inequality like \( \frac{4(x+2)}{3+2x} \leq 5 \) is converted into a proper rational inequality by handling the fractions skillfully.

Interval Notation

Interval notation is a shorthand way of representing a set of numbers along a number line. It uses parentheses and brackets to show which endpoints are included or excluded in the interval.
Here's a quick guide on how to use interval notation effectively:

• Round brackets ( ) indicate that the endpoint is not included.
• Square brackets [ ] indicate that the endpoint is included.
• For example, the interval \([-3, 5)\) includes -3 but does not include 5.
• The interval \( (-\frac{7}{6}, +\backslash\text{infinity}) \) includes all numbers greater than or equal to \(-\frac{7}{6} \) but never reaches infinity.

Use this concise notation to express the solution set of your rational inequalities succinctly.

Inequality Solving Steps

To efficiently solve a rational inequality, follow these structured steps:

• Step 1: Simplify the Inequality. Rewrite the inequality in a simplified form.
• Step 2: Eliminate the Fraction. If an inequality contains fractions, multiply both sides by the denominator to eliminate them.
• Step 3: Distribute and Combine Like Terms. Simplify by distributing and combining like terms.
• Step 4: Solve for the Variable. Isolate the variable by performing basic algebraic operations.
• Step 5: Consider the Denominator's Restriction. Ensure the solution does not include values that make the denominator zero.
• Step 6: Combine Conditions. Merge all conditions to write the final solution set in interval notation.

These steps are crucial in breaking down and solving rational inequalities with precision.

Eliminating Fractions in Inequalities

Eliminating fractions in an inequality is necessary for easier solving. Here’s how to do it effectively:
Consider an example inequality \(\frac{4(x+2)}{3+2x} \leq 5\).
To eliminate the fraction:

• Multiply both sides by the denominator \( (3+2x) \) to remove the fraction. This results in \ 4(x + 2) \leq 5(3 + 2x) \.
• Be cautious: If the denominator can be negative, remember the inequality sign flips when multiplied by a negative value.
• After removing the fraction, continue simplifying by distributing and combining like terms to isolate the variable.

By eliminating fractions, the inequality becomes simpler and ready for further algebraic manipulation.