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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 38a

Exercises 27–40 contain linear equations with constants in denominators. Solve each equation. 5 + (x - 2)/3 = (x + 3)/8

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Step 1: Identify the least common denominator (LCD) of the fractions in the equation. The denominators are 3 and 8, so the LCD is 24.
Step 2: Multiply every term in the equation by the LCD (24) to eliminate the fractions. This gives: 24 * 5 + 24 * (x - 2)/3 = 24 * (x + 3)/8.
Step 3: Simplify each term after multiplying by the LCD. For example, 24 * (x - 2)/3 becomes 8 * (x - 2), and 24 * (x + 3)/8 becomes 3 * (x + 3). Rewrite the equation as: 120 + 8(x - 2) = 3(x + 3).
Step 4: Distribute the constants across the parentheses. For example, 8(x - 2) becomes 8x - 16, and 3(x + 3) becomes 3x + 9. Rewrite the equation as: 120 + 8x - 16 = 3x + 9.
Step 5: Combine like terms and isolate the variable x. Combine constants on one side and x terms on the other side to solve for x.

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