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$x+y=12$

$y=x^{2}$

If $(x,y)$ is a solution to the system of equations above, which of the following is a possible value of $x$?

","hasPart":{"@type":"Question","typicalAgeRange":"12-22","educationalAlignment":[{"@type":"AlignmentObject","alignmentType":"educationalSubject","targetName":"SAT"},{"@type":"AlignmentObject","alignmentType":"educationalLevel","targetName":"SAT","educationalFramework":"Common Core","targetUrl":"https://classnotes.xyz/mcqs/sat-math-advanced-math-multiple-choice-question-107"}],"about":{"@type":"Thing","name":"Math, Advanced Math, Duplicate"},"educationalLevel":"intermediate","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","assesses":["Math","Advanced Math","Duplicate"],"name":"

$x+y=12$

$y=x^{2}$

If $(x,y)$ is a solution to the system of equations above, which of the following is a possible value of $x$?

","comment":{"@type":"Comment","text":"

Choice D is correct. Substituting $\\bm{x}^{2}$ from the second equation for $y$ in the first equation yields $x+\\bm{x}^{2}=12$. Subtracting 12 from both sides of this equation and rewriting the equation results in $x^{2}+\\bm{x}-12=0$. Factoring the left-hand side of this equation yields $(\\bm{x}-3)(\\bm{x}+4)=0$. Using the zero product property to solve for x, it follows that $\\bm{x}-3=0$ and $\\bm{x}+4=0$. Solving each equation for $x$ yields $\\bm{x}=\\bm{3}$ and $\\bm{x}=-4$, respectively. Thus, two possible values of $x$ are $3$ and $-4$. Of the choices given, $3$ is the only possible value of $x$.

Choices A, B, and C are incorrect. Substituting 0 for $x$ in the first equation gives $0+\\bm{y}=12$, or $\\bm{y}=\\bm{1}2$; then, substituting 12 for $y$ and 0 for $x$ in the second equation gives $12=0^{2}$, or $12=0$, which is false. Similarly, substituting 1 or 2 for $x$ in the first equation yields $\\bm{y}=11$ or $\\bm{y}=10$, respectively; then, substituting 11 or 10 for $y$ in the second equation yields a false statement.

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$x+y=12$

$y=x^{2}$

If $(x,y)$ is a solution to the system of equations above, which of the following is a possible value of $x$?

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$x+y=12$$y=x^{2}$If $(x,y)$ is a solution to the system of equations above, which of the following is a possible value of $x$?

$x+y=12$$y=x^{2}$If $(x,y)$ is a solution to the system of equations above, which of the following is a possible value of $x$?

$x+y=12$$y=x^{2}$If $(x,y)$ is a solution to the system of equations above, which of the following is a possible value of $x$?

$x+y=12$

$y=x^{2}$

If $(x,y)$ is a solution to the system of equations above, which of the following is a possible value of $x$?

$x+y=12$

$y=x^{2}$

If $(x,y)$ is a solution to the system of equations above, which of the following is a possible value of $x$?

$x+y=12$

$y=x^{2}$

If $(x,y)$ is a solution to the system of equations above, which of the following is a possible value of $x$?