Table of contents
- 0. Review of Algebra4h 18m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations1h 43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 5m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 22m
- 10. Combinatorics & Probability1h 45m
1. Equations & Inequalities
Linear Equations
Problem 53a
Textbook Question
Exercises 41–60 contain rational equations with variables in denominators. For each equation, a. write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. 3/(2x - 2) + 1/2 = 2/(x - 1)
Verified step by step guidance1
Identify the denominators in the equation: \(2x - 2\), \(2\), and \(x - 1\).
Determine the restrictions on the variable by setting each denominator equal to zero and solving for \(x\): \(2x - 2 = 0\) gives \(x = 1\), and \(x - 1 = 0\) gives \(x = 1\). Therefore, \(x = 1\) is a restriction because it makes the denominators undefined.
Rewrite the equation: \(\frac{3}{2x - 2} + \frac{1}{2} = \frac{2}{x - 1}\). Notice that \(2x - 2\) can be factored as \(2(x - 1)\), so rewrite the first term as \(\frac{3}{2(x - 1)}\). The equation becomes \(\frac{3}{2(x - 1)} + \frac{1}{2} = \frac{2}{x - 1}\).
Eliminate the fractions by multiplying through by the least common denominator (LCD), which is \(2(x - 1)\). Multiply each term by \(2(x - 1)\): \(2(x - 1) \cdot \frac{3}{2(x - 1)} + 2(x - 1) \cdot \frac{1}{2} = 2(x - 1) \cdot \frac{2}{x - 1}\).
Simplify each term: The first term simplifies to \(3\), the second term simplifies to \((x - 1)\), and the third term simplifies to \(4\). The resulting equation is \(3 + (x - 1) = 4\). Solve this linear equation for \(x\), keeping in mind the restriction \(x \neq 1\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Rational Equations
Rational equations are equations that involve fractions with polynomials in the numerator and denominator. To solve these equations, it is essential to find a common denominator and eliminate the fractions, which simplifies the equation. Understanding how to manipulate these fractions is crucial for solving rational equations effectively.
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Restrictions on Variables
Restrictions on variables in rational equations arise when the denominator equals zero, as division by zero is undefined. Identifying these restrictions is critical because they determine the values that the variable cannot take. For example, in the equation given, setting the denominator to zero helps find the values of x that must be excluded from the solution set.
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Equations with Two Variables
Solving for Variables
Solving for variables in rational equations involves isolating the variable on one side of the equation after addressing any restrictions. This process may include cross-multiplication, combining like terms, and applying inverse operations. It is important to check the final solutions against the identified restrictions to ensure they are valid.
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Related Practice
Textbook Question
Solve each equation for x. 2(x-a) +b =3x+a
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