What is the solution to the equation above?

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 multipliion 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 isn’t a solution to the given equation. Substituting 3 and 5 into the given equation yields similarly false statements.

Question what-i

What is the solution to the equation above?

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