Textbook Question
In Exercises 107–110, use graphs to find each set. (-2,1] ∩ [-1,3)
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Ch. 1 - Equations and Inequalities
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
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Table of contents
- Ch. P - Fundamental Concepts of Algebra311
- Ch. 1 - Equations and Inequalities398
- Ch. 2 - Functions and Graphs407
- Ch. 3 - Polynomial and Rational Functions306
- Ch. 4 - Exponential and Logarithmic Functions247
- Ch. 5 - Systems of Equations and Inequalities173
- Ch. 6 - Matrices and Determinants143
- Ch. 7 - Conic Sections124
- Ch. 8 - Sequences, Induction, and Probability251
Chapter 2, Problem 106
Solve each equation in Exercises 83–108 by the method of your choice. 1/x + 1/(x + 3) = 1/4
Verified step by step guidance1
Identify the equation: \(\frac{1}{x} + \frac{1}{x + 3} = \frac{1}{4}\).
Find the least common denominator (LCD) for the fractions, which is \(4x(x + 3)\).
Multiply both sides of the equation by the LCD to eliminate the denominators: \(4x(x + 3) \times \left( \frac{1}{x} + \frac{1}{x + 3} \right) = 4x(x + 3) \times \frac{1}{4}\).
Simplify each term after multiplication: \(4(x + 3) + 4x = x(x + 3)\).
Rewrite the equation and expand all terms to form a quadratic equation: \(4x + 12 + 4x = x^2 + 3x\), then combine like terms and set the equation to zero.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Solving Rational Equations
Rational equations involve expressions with variables in the denominator. To solve them, find a common denominator to combine terms or clear denominators by multiplying both sides, ensuring to check for excluded values that make denominators zero.
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Finding the Least Common Denominator (LCD)
The LCD is the smallest expression that all denominators divide into evenly. Identifying the LCD allows you to combine fractions or eliminate denominators by multiplying through, simplifying the equation to a polynomial or linear form.
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03:42Rationalizing Denominators Using Conjugates
Checking for Extraneous Solutions
When solving rational equations, some solutions may make denominators zero, which are invalid. After finding potential solutions, substitute them back into the original equation to ensure they do not cause division by zero.
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Related Practice
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Textbook Question
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Textbook Question
Solve each equation in Exercises 83–108 by the method of your choice. 1/x + 1/(x + 2) = 1/3
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