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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 33a

Exercises 27–40 contain linear equations with constants in denominators. Solve each equation. 3x/5 - x = x/10 - 5/2

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Combine all terms involving x on one side of the equation. Start by subtracting \( \frac{x}{10} \) from both sides: \( \frac{3x}{5} - x - \frac{x}{10} = -\frac{5}{2} \).
Find a common denominator for the fractions involving x on the left-hand side. The least common denominator (LCD) of 5, 1, and 10 is 10. Rewrite each term with this denominator: \( \frac{6x}{10} - \frac{10x}{10} - \frac{x}{10} = -\frac{5}{2} \).
Simplify the terms on the left-hand side by combining the fractions: \( \frac{6x - 10x - x}{10} = -\frac{5}{2} \), which simplifies further to \( \frac{-5x}{10} = -\frac{5}{2} \).
Eliminate the fractions by multiplying through by the least common denominator of the denominators on both sides. The LCD of 10 and 2 is 10. Multiply through by 10: \( 10 \cdot \frac{-5x}{10} = 10 \cdot -\frac{5}{2} \), resulting in \( -5x = -25 \).
Solve for x by dividing both sides of the equation by -5: \( x = \frac{-25}{-5} \). Simplify the result to find the value of x.

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