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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 2, Problem 53a

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)

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1
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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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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