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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 1 - Equations and InequalitiesProblem 65a
Chapter 2, Problem 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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Rational Functions

Rational functions are expressions formed by the ratio of two polynomials. In this question, y1, y2, and y3 are rational functions where the numerator and denominator are polynomials. Understanding how to manipulate and combine these functions is essential for solving the equation y1 + y2 = y3.
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Finding Common Denominators

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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Solving Polynomial Equations

Once the rational functions are combined, the resulting equation will typically be a polynomial equation. Solving polynomial equations involves finding the values of x that satisfy the equation, which may require factoring, using the quadratic formula, or applying other algebraic techniques. This is the final step in determining the values of x that meet the given conditions.
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Related Practice
Textbook Question

Find all values of x satisfying the given conditions. y1 = (2x - 1)/(x2 + 2x - 8), y2 = 2/(x + 4), y3 = 1/(x - 2), and y1 + y2 = y3.

Textbook Question

Solve each equation in Exercises 65–74 using the quadratic formula. x2+8x+15=0x^2 + 8x + 15 = 0

Textbook Question

Perform the indicated operation(s) and write the result in standard form. (2 - 3i)(1 - i) - (3 - i)(3 + i)

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Textbook Question

In Exercises 65–70, perform the indicated operation(s) and write the result in standard form. (2 + i)2 - (3 - i)2

Textbook Question

In Exercises 61–76, solve each absolute value equation or indicate that the equation has no solution. |2x - 1| = 5

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Textbook Question

Perform the indicated operation(s) and write the result in standard form. (8 + 9i)(2 - i) - (1 - i)(1 + i)