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

$y equals, x squared plus 2 x plus 1, and, x plus y plus 1, equals 0$

If $the ordered pair x subscript 1 comma y subscript 1$ and $the ordered pair x subscript 2 comma y subscript 2$ are the two solutions to the system of equations above, what is the value of $y subscript 1, plus, y subscript 2$ ?

A. $negative 3$

B. $negative 2$

C. $negative 1$

D. $1$

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Explanation

Choice D is correct. The system of equations can be solved using the substitution method. Solving the second equation for y gives y = –x – 1. Substituting the expression –x – 1 for y into the first equation gives –x – 1 = x² + 2x + 1. Adding x + 1 to both sides of the equation yields x² + 3x + 2 = 0. The left-hand side of the equation can be factored by finding two numbers whose sum is 3 and whose product is 2, which gives (x + 2)(x + 1) = 0. Setting each factor equal to 0 yields x + 2 = 0 and x + 1 = 0, and solving for x yields x = –2 or x = –1. These values of x can be substituted for x in the equation y = –x – 1 to find the corresponding y-values: y = –(–2) – 1 = 2 – 1 = 1 and y = –(–1) – 1 = 1 – 1 = 0. It follows that (–2, 1) and (–1, 0) are the solutions to the given system of equations. Therefore, (x₁, y₁) = (–2, 1), (x₂, y₂) = (–1, 0), and y₁ + y₂ = 1 + 0 = 1.

Choice A is incorrect. The solutions to the system of equations are (x₁, y₁) = (–2, 1) and (x₂, y₂) = (–1, 0). Therefore, –3 is the sum of the x-coordinates of the solutions, not the sum of the y-coordinates of the solutions. Choices B and C are incorrect and may be the result of computation or substitution errors.