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

the fraction with numerator 2 times, open parenthesis, x plus 1, close parenthesis, and denominator x plus 5, end fraction, equals 1 minus the fraction with numerator 1, and denominator x plus 5, end fraction

What is the solution to the equation above?

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Explanation

Choice B is correct. Since the fraction with numerator x plus 5, and denominator x plus 5, end fraction is equivalent to 1, the right-hand side of the given equation can be rewritten as the fraction with numerator x plus 5, and denominator x plus 5, end fraction, minus, the fraction 1 over, x plus 5, end fraction , or the fraction with numerator x plus 4, and denominator x plus 5, end fraction. Since the left- and right-hand sides of the equation the fraction with numerator 2 times, open parenthesis, x plus 1, close parenthesis, and denominator x plus 5, end fraction, equals, the fraction with numerator x plus 4, and denominator x plus 5, end fraction have the same denominator, it follows that 2 times, open parenthesis, x plus 1, close parenthesis, equals x plus 4. Applying the distributive property of multiplication to the expression 2 times, open parenthesis, x plus 1, close parenthesis yields 2 times x, plus, 2 times 1, or 2 x plus 2. Therefore, 2 x plus 2, equals, x plus 4. Subtracting x and 2 from both sides of this equation yields x equals 2.

Choices A, C, and D are incorrect. If x equals 0, then the fraction with numerator 2 times, open parenthesis, 0 plus 1, close parenthesis, and denominator 0 plus 5, end fraction, equals 1 minus, the fraction with numerator 1, and denominator 0 plus 5, end fraction. This can be rewritten as 2 over 5, equals 4 over 5, which is a false statement. Therefore, 0 isnโ€™t a solution to the given equation. Substituting 3 and 5 into the given equation yields similarly false statements.