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9. Anne was given the following problem: r+2 (+2)(x+3) エ十3 When solving, she found the answers to be x =-2 and x =4. Are both answers reasonable solutions to this problem? Explain your reasoning. G. x+2 z+3

9. Anne was given the following problem: r+2 (+2)(x+3) エ十3 When solving, she found the answers to be x =-2 and x =4. Are both answers reasonable solutions to this problem? Explain your reasoning. G. x+2 z+3

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Certainly! Here's the transcription and explanation: --- ### Problem 9 Anne was given the following problem: \[ \frac{x}{x+2} + \frac{2}{(x+2)(x+3)} = \frac{5}{x+3} \] When solving, she found the answers to be \(x = -2\) and \(x = 4\). Are both answers reasonable solutions to this problem? Explain your reasoning. **Explanation:** The given equation involves rational expressions with variables in the denominator. When solving such equations, it is crucial to identify any restrictions on the variable \(x\), which occur when the denominators are equal to zero. - The expression \(\frac{x}{x+2}\) implies that \(x \neq -2\). - The expression \(\frac{2}{(x+2)(x+3)}\) implies that \(x \neq -2\) and \(x \neq -3\). - The expression \(\frac{5}{x+3}\) implies that \(x \neq -3\). Thus, any solution that results in division by zero must be excluded. Given Anne's solutions: - \(x = -2\) is invalid since it causes the denominators \(x + 2\) and \((x+2)(x+3)\) to become zero. - \(x = 4\) does not result in any denominators being zero and thus is a valid solution. Thus, only \(x = 4\) is a reasonable solution.
Transcribed Image Text:Certainly! Here's the transcription and explanation: --- ### Problem 9 Anne was given the following problem: \[ \frac{x}{x+2} + \frac{2}{(x+2)(x+3)} = \frac{5}{x+3} \] When solving, she found the answers to be \(x = -2\) and \(x = 4\). Are both answers reasonable solutions to this problem? Explain your reasoning. **Explanation:** The given equation involves rational expressions with variables in the denominator. When solving such equations, it is crucial to identify any restrictions on the variable \(x\), which occur when the denominators are equal to zero. - The expression \(\frac{x}{x+2}\) implies that \(x \neq -2\). - The expression \(\frac{2}{(x+2)(x+3)}\) implies that \(x \neq -2\) and \(x \neq -3\). - The expression \(\frac{5}{x+3}\) implies that \(x \neq -3\). Thus, any solution that results in division by zero must be excluded. Given Anne's solutions: - \(x = -2\) is invalid since it causes the denominators \(x + 2\) and \((x+2)(x+3)\) to become zero. - \(x = 4\) does not result in any denominators being zero and thus is a valid solution. Thus, only \(x = 4\) is a reasonable solution.
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