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Advanced Math / Nonlinear equations in one variable and systems of equations in two variables Difficulty: Hard

$the fraction with numerator 4 x squared, and denominator x squared minus 9, end fraction, minus the fraction with numerator 2 x, and denominator x plus 3, end fraction, equals the fraction with numerator 1, and denominator x minus 3, end fraction$

What value of x satisfies the equation above?

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Explanation

Choice C is correct. Each fraction in the given equation can be expressed with the common denominator $x squared, minus 9$ . Multiplying $the fraction with numerator 2 x, and denominator x plus 3, end fraction$ by yields $the fraction with numerator 2 x squared, minus 6, and denominator x squared minus 9, end fraction$ , and multiplying $the fraction with numerator 1, and denominator x minus 3, end fraction$ by $the fraction with numerator x plus 3, and denominator x plus 3, end fraction$ yields $the fraction with numerator x plus 3, and denominator x squared minus 9, end fraction$ . Therefore, the given equation can be written as $the fraction with numerator 4 x squared, and denominator x squared minus 9, minus, the fraction with numerator 2 x squared, minus 6 x, and denominator x squared minus 9, end fraction, equals, the fraction with numerator x plus 3, and denominator x squared minus 9, end fraction$ . Multiplying each fraction by the denominator results in the equation $4 x squared minus, open parenthesis, 2 x squared, minus 6 x, close parenthesis, equals, x plus 3$ , or $2 x squared, plus 6 x, equals, x plus 3$ . This equation can be solved by setting a quadratic expression equal to 0, then solving for x. Subtracting $x plus 3$ from both sides of this equation yields $2 x squared, plus 5 x, minus 3, equals 0$ . The expression $2 x squared, plus 5 x, minus 3$ can be factored, resulting in the equation $open parenthesis, 2 x minus 1, close parenthesis, times, open parenthesis, x plus 3, close parenthesis, equals 0$ . By the zero product property, $2 x minus 1, equals 0$ or $x plus 3, equals 0$ . To solve for x in $2 x minus 1, equals 0$ , 1 can be added to both sides of the equation, resulting in $2 x equals 1$ . Dividing both sides of this equation by 2 results in $x equals one half$ . Solving for x in $x plus 3, equals 0$ yields $x equals negative 3$ . However, this value of x would result in the second fraction of the original equation having a denominator of 0. Therefore, $x equals negative 3$ is an extraneous solution. Thus, the only value of x that satisfies the given equation is $x equals one half$ .

Choice A is incorrect and may result from solving $x plus 3, equals 0$ but not realizing that this solution is extraneous because it would result in a denominator of 0 in the second fraction. Choice B is incorrect and may result from a sign error when solving $2 x minus 1, equals 0$ for x. Choice D is incorrect and may result from a calculation error.