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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 44

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. 7/2x - 5/3x = 22/3

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Identify the denominators in the equation \(\frac{7}{2x} - \frac{5}{3x} = \frac{22}{3}\). The denominators are \$2x$ and \$3x$.
Find the values of \(x\) that make any denominator zero. Set each denominator equal to zero: \(2x = 0\) and \(3x = 0\). Solve these to find the restrictions on \(x\).
Rewrite the equation to have a common denominator on the left side. The common denominator for \$2x$ and \$3x$ is \$6x$. Express each fraction with denominator \$6x$:
\(\frac{7}{2x} = \frac{7 \times 3}{6x} = \frac{21}{6x}\) and \(\frac{5}{3x} = \frac{5 \times 2}{6x} = \frac{10}{6x}\).
Combine the fractions on the left side: \(\frac{21}{6x} - \frac{10}{6x} = \frac{21 - 10}{6x} = \frac{11}{6x}\). So the equation becomes \(\frac{11}{6x} = \frac{22}{3}\).
Solve for \(x\) by cross-multiplying: \(11 \times 3 = 22 \times 6x\). Simplify and solve the resulting equation for \(x\), keeping in mind the restrictions found earlier.

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