Find all values of x satisfying the given conditions. y1 = 2x/(x + 2), y2 = 3/(x + 4), and y1 + y2 = 1

Ch. 1 - Equations and Inequalities

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Blitzer 8th Edition

Ch. 1 - Equations and Inequalities

Problema 119

Capitolo 2, Problema 119

Find all values of x satisfying the given conditions. y₁ = 2x/(x + 2), y₂ = 3/(x + 4), and y₁ + y₂ = 1

Guida verificata passo dopo passo

Step 1: Start by substituting the given expressions for y1 and y2 into the equation y1 + y2 = 1. This gives: \( \frac{2x}{x + 2} + \frac{3}{x + 4} = 1 \).

Step 2: To simplify the equation, find a common denominator for the fractions. The denominators are \( x + 2 \) and \( x + 4 \), so the common denominator is \( (x + 2)(x + 4) \). Rewrite each fraction with this common denominator.

Step 3: Combine the fractions into a single fraction. This results in: \( \frac{2x(x + 4) + 3(x + 2)}{(x + 2)(x + 4)} = 1 \).

Step 4: Multiply through by the common denominator \( (x + 2)(x + 4) \) to eliminate the fractions. This gives: \( 2x(x + 4) + 3(x + 2) = (x + 2)(x + 4) \).

Step 5: Expand all terms and simplify the equation. Combine like terms and rearrange the equation into standard form (e.g., \( ax^2 + bx + c = 0 \)) to solve for \( x \).

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Rational Functions

Rational functions are expressions formed by the ratio of two polynomials. In this question, y1 and y2 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 = 1.

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Intro to Rational Functions

Finding Common Denominators

To add or equate rational functions, it is often necessary to find a common denominator. This involves identifying the least common multiple of the denominators of the functions involved. In this case, combining y1 and y2 requires finding a common denominator to simplify the equation effectively.

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Solving Rational Equations

Solving rational equations involves isolating the variable, often by eliminating the denominators through multiplication. This process may introduce extraneous solutions, so it is crucial to check each solution against the original equations. In this problem, solving the equation y1 + y2 = 1 will require careful algebraic manipulation to find valid values of x.

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