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Algebra / Systems of two linear equations in two variables Difficulty: Hard

If a new graph of three linear equations is created using the system of equations shown and the equation $x + 4 y = - 16$ , how many solutions $(x, y)$ will the resulting system of three equations have?

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Explanation

Choice A is correct. A solution to a system of equations must satisfy each equation in the system. It follows that if an ordered pair $(x, y)$ is a solution to the system, the point $(x, y)$ lies on the graph in the xy-plane of each equation in the system. The only point that lies on each graph of the system of two linear equations shown is their intersection point $left parenthesis 8 comma 2 right parenthesis$ . It follows that if a new graph of three linear equations is created using the system of equations shown and the graph of $x + 4 y = - 16$ , this system has either zero solutions or one solution, the point $left parenthesis 8 comma 2 right parenthesis$ . Substituting $8$ for $x$ and $2$ for $y$ in the equation $x + 4 y = - 16$ yields $8 + 4 (2) = - 16$ , or $16 = - 16$ . Since this equation is not true, the point $left parenthesis 8 comma 2 right parenthesis$ does not lie on the graph of $x + 4 y = - 16$ . Therefore, $left parenthesis 8 comma 2 right parenthesis$ is not a solution to the system of three equations. It follows that there are zero solutions to this system.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.