Table of contents
- 0. Review of Algebra4h 16m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 6m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 19m
- 10. Combinatorics & Probability1h 45m
1. Equations & Inequalities
Rational Equations
Problem 17b
Textbook Question
Solve each equation. | 6x + 1/ x - 1 | = 3

1
Identify the absolute value equation: \(|\frac{6x + 1}{x - 1}| = 3\).
Set up two separate equations to account for the absolute value: \(\frac{6x + 1}{x - 1} = 3\) and \(\frac{6x + 1}{x - 1} = -3\).
Solve the first equation: \(\frac{6x + 1}{x - 1} = 3\). Multiply both sides by \(x - 1\) to eliminate the fraction, resulting in \(6x + 1 = 3(x - 1)\).
Solve the second equation: \(\frac{6x + 1}{x - 1} = -3\). Similarly, multiply both sides by \(x - 1\) to eliminate the fraction, resulting in \(6x + 1 = -3(x - 1)\).
Solve each resulting linear equation for \(x\) and check for any extraneous solutions by substituting back into the original equation.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Absolute Value
Absolute value measures the distance of a number from zero on the number line, disregarding its sign. In the equation |6x + 1/(x - 1)| = 3, the absolute value indicates that the expression inside can equal either 3 or -3. Understanding how to manipulate absolute values is crucial for solving equations that involve them.
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Rational Expressions
A rational expression is a fraction where both the numerator and the denominator are polynomials. In this equation, 1/(x - 1) is a rational expression, and it is important to consider its domain, as it cannot equal zero. This understanding helps in identifying restrictions on the variable x when solving the equation.
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Equating Expressions
When solving equations, particularly those involving absolute values, it is essential to set up separate equations based on the possible cases. For |A| = B, we create A = B and A = -B. This method allows us to find all potential solutions, which we can then verify to ensure they satisfy the original equation.
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