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

$Equation 1: x minus y, equals 1. Equation 2: x plus y equals, x squared minus 3$

Which ordered pair is a solution to the system of equations above?

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Explanation

Choice A is correct. The solution to the given system of equations can be found by solving the first equation for x, which gives $x equals, y plus 1$ , and substituting that value of x into the second equation which gives $y plus 1, plus y, equals, open parenthesis, y plus 1, close parenthesis, squared, minus 3$ . Rewriting this equation by adding like terms and expanding $open parenthesis, y plus 1, close parenthesis, squared$ gives $2 y plus 1, equals, y squared plus 2 y, minus 2$ . Subtracting $2 y$ from both sides of this equation gives $1 equals, y squared minus 2$ . Adding to 2 to both sides of this equation gives $3 equals y squared$ . Therefore, it follows that $y equals, plus or minus the square root of 3$ . Substituting $the square root of 3$ for y in the first equation yields $x minus the square root of 3, equals 1$ . Adding $the square root of 3$ to both sides of this equation yields $x equals, 1 plus the square root of 3$ . Therefore, the ordered pair $1 plus the square root of 3, end root, comma the square root of 3$ is a solution to the given system of equations.

Choice B is incorrect. Substituting $the square root of 3$ for x and $the negative of the square root of 3$ for y in the first equation yields $the square root of 3, end root, minus, the negative of the square root of 3, equals 1$ , or $2 times the square root of 3, equals 1$ , which isn’t a true statement. Choice C is incorrect. Substituting $1 plus the square root of 5$ for x and $the square root of 5$ for y in the second equation yields $open parenthesis, 1 plus the square root of 5, close parenthesis, plus the square root of 5, equals, open parenthesis, 1 plus the square root of 5, close parenthesis, squared, minus 3$ , or $1 plus, 2 times the square root of 5, equals, 2 times the square root of 5, plus 3$ , which isn’t a true statement. Choice D is incorrect. Substituting $the square root of 5$ for x and $open parenthesis, negative 1 plus the square root of 5, close parenthesis$ for y in the second equation yields $the square root of 5 plus, open parenthesis, negative 1 plus the square root of 5, close parenthesis, equals, the square root of 5, squared, minus 3$ , or $2 times the square root of 5, minus 1, equals 2$ , which isn’t a true statement.