Textbook Question
Graph each equation in Exercises 13 - 28. Let x = - 3, - 2, - 1, 0, 1, 2, 3
y = x3 - 1
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 27
Exercises 27–40 contain linear equations with constants in denominators. Solve each equation. x/3 = x/2 - 2
Verified step by step guidance1
Identify the equation: \(\frac{x}{3} = \frac{x}{2} - 2\).
Find the least common denominator (LCD) of the denominators 3 and 2, which is 6.
Multiply every term in the equation by the LCD (6) to eliminate the denominators: \(6 \times \frac{x}{3} = 6 \times \frac{x}{2} - 6 \times 2\).
Simplify each term after multiplication: \(2x = 3x - 12\).
Isolate the variable \(x\) by subtracting \$3x\( from both sides: \(2x - 3x = -12\), then simplify to get \)-x = -12$.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Solving Linear Equations
A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. Solving such equations involves isolating the variable on one side to find its value. Techniques include combining like terms and performing inverse operations.
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Solving Linear Equations with Fractions
Clearing Fractions by Multiplying Both Sides
When an equation contains fractions, multiplying both sides by the least common denominator (LCD) eliminates the denominators, simplifying the equation. This step helps avoid dealing with fractions and makes solving the equation more straightforward.
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Checking for Extraneous Solutions
After solving an equation, especially those involving variables in denominators, it is important to check solutions by substituting them back into the original equation. This ensures no solution makes a denominator zero, which would be undefined and thus extraneous.
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Related Practice
Textbook Question
The length of a rectangular pool is 6 meters less than twice the width. If the pool's perimeter is 126 meters, what are its dimensions?
Textbook Question
In Exercises 21–28, divide and express the result in standard form. (2 + 3i)/(2 + i)
Textbook Question
The length of the rectangular tennis court at Wimbledon is 6 feet longer than twice the width. If the court's perimeter is 228 feet, what are the court's dimensions?
Textbook Question
Graph each equation in Exercises 13 - 28. Let x = - 3, - 2, - 1, 0, 1, 2, 3
y = x3
Textbook Question
In all exercises, other than exercises with no solution, use interval notation to express solution sets and graph each solution set on a number line. In Exercises 27–50, solve each linear inequality. 5x + 11 < 26