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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 69

In Exercises 67–70, find all values of x such that y = 0. y=x+63x−12−5x−4−23y = \(\frac{x + 6}{3x - 12}\) - \(\frac{5}{x - 4}\) - \(\frac{2}{3}\)

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Start with the given equation: \(y = \frac{x + 6}{3x - 12} - \frac{5}{x - 4} - \frac{2}{3}\). We want to find all values of \(x\) such that \(y = 0\), so set the equation equal to zero: \(\frac{x + 6}{3x - 12} - \frac{5}{x - 4} - \frac{2}{3} = 0\).
Notice that \(3x - 12\) can be factored as \(3(x - 4)\). Rewrite the equation using this factorization: \(\frac{x + 6}{3(x - 4)} - \frac{5}{x - 4} - \frac{2}{3} = 0\).
To combine the fractions, find the least common denominator (LCD). The denominators are \(3(x - 4)\), \(x - 4\), and \(3\). The LCD is \(3(x - 4)\). Rewrite each term with this common denominator:
\(\frac{x + 6}{3(x - 4)} - \frac{5 \cdot 3}{3(x - 4)} - \frac{2(x - 4)}{3(x - 4)} = 0\).
Combine the numerators over the common denominator:
\(\frac{(x + 6) - 15 - 2(x - 4)}{3(x - 4)} = 0\).
Since a fraction equals zero only when its numerator is zero (and the denominator is not zero), set the numerator equal to zero and solve for \(x\): \((x + 6) - 15 - 2(x - 4) = 0\). Then simplify and solve the resulting linear equation.

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