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Chapter 2: Problem 97

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I began the solution of the rational inequality \(\frac{x+1}{x+3} \geq 2\) by setting both \(x+1\) and \(x+3\) equal to zero.

Short Answer

Expert verified
The statement does not make sense because setting the numerator equal to zero is not a necessary step in solving this rational inequality. The correct first step would be to rearrange the inequality to have zero on one side and then solve the resulting quadratic inequality.

Step by step solution

01

Understand the Statement

The given statement says 'I began the solution of the rational inequality \(\frac{x+1}{x+3} \geq 2\) by setting both \(x+1\) and \(x+3\) equal to zero.' This implies that each of these separate terms (the numerator \(x+1\) and the denominator \(x+3\)) is being set equal to zero and presumably solved for \(x\). This is done typically in algebra to find the roots of the equation or points of discontinuity.
02

Evaluate the Validity of the Approach

The first step to solving the inequality \(\frac{x+1}{x+3} \geq 2\) should not be setting \(x+1\) and \(x+3\) equal to zero. Although it's true that setting the denominator equal to zero helps in identifying values that cause the fraction to be undefined and thus excluded from possible solutions, setting the numerator equal to zero is not a necessary step in the solution process for this particular inequality. Instead, the rational inequality should be rearranged to have zero on one side and then solving the resulting quadratic inequality.
03

Clarify the Correct Approach

The correct step in solving the rational inequality \(\frac{x+1}{x+3} \geq 2\) would involve first subtracting 2 from both sides to create a standard quadratic inequality, i.e., \(\frac{x+1}{x+3} - 2 \geq 0\). This will result in \(\frac{x+1- 2*(x+3)}{x+3} \geq 0\) or \(\frac{-x-5}{x+3} \geq 0\). Then finding the zeroes of this ratio and the points of discontinuity, and creating a sign chart or performing other appropriate solution methods for quadratic inequalities can be conducted.

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