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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 65a

In Exercises 61–66, find all values of x satisfying the given conditions. y1 = 5/(x + 4), y2 = 3/(x + 3), y3 = (12x + 19)/(x2 + 7x + 12). and y1 + y2 = y3.

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Start by writing the given equation y1 + y2 = y3 explicitly in terms of x: \( \frac{5}{x + 4} + \frac{3}{x + 3} = \frac{12x + 19}{x^2 + 7x + 12} \).
Factor the denominator \( x^2 + 7x + 12 \) on the right-hand side. This quadratic can be factored as \( (x + 3)(x + 4) \), so the equation becomes \( \frac{5}{x + 4} + \frac{3}{x + 3} = \frac{12x + 19}{(x + 3)(x + 4)} \).
Combine the fractions on the left-hand side over the common denominator \( (x + 3)(x + 4) \). This gives \( \frac{5(x + 3) + 3(x + 4)}{(x + 3)(x + 4)} \). Simplify the numerator: \( 5(x + 3) + 3(x + 4) = 5x + 15 + 3x + 12 = 8x + 27 \).
Set the simplified left-hand side equal to the right-hand side: \( \frac{8x + 27}{(x + 3)(x + 4)} = \frac{12x + 19}{(x + 3)(x + 4)} \). Since the denominators are the same, equate the numerators: \( 8x + 27 = 12x + 19 \).
Solve the equation \( 8x + 27 = 12x + 19 \) for \( x \). Subtract \( 8x \) from both sides: \( 27 = 4x + 19 \). Subtract 19 from both sides: \( 8 = 4x \). Finally, divide both sides by 4: \( x = 2 \). Verify that \( x = 2 \) does not make any denominator zero (it does not).

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To add or equate rational functions, it is often necessary to find a common denominator. This involves identifying a common multiple of the denominators of the functions involved, which allows for the combination of the fractions into a single expression. This step is crucial for simplifying the equation and solving for x.
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