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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 31

Exercises 27–40 contain linear equations with constants in denominators. Solve each equation. 3x/5 = 2x/3 + 1

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Identify the given equation: \(\frac{3x}{5} = \frac{2x}{3} + 1\).
Find the least common denominator (LCD) of the fractions involved. Here, the denominators are 5 and 3, so the LCD is 15.
Multiply every term on both sides of the equation by the LCD (15) to eliminate the denominators: \(15 \times \frac{3x}{5} = 15 \times \frac{2x}{3} + 15 \times 1\).
Simplify each term after multiplication: \(15 \times \frac{3x}{5} = 3 \times 3x = 9x\), \(15 \times \frac{2x}{3} = 5 \times 2x = 10x\), and \(15 \times 1 = 15\).
Rewrite the equation without fractions: \(9x = 10x + 15\). Then, solve for \(x\) by isolating the variable on one side.

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