{"@context":"https://schema.org/","@type":"Quiz","name":"

$x-y=1$

$x+y=x^{2}-3$

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

","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-10"}],"about":{"@type":"Thing","name":"Math, Advanced Math, Duplicate"},"educationalLevel":"intermediate","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","assesses":["Math","Advanced Math","Duplicate"],"name":"

$x-y=1$

$x+y=x^{2}-3$

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

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

Choice A is correct. The solution to the given system of equations can be found by solving the first equation for $x$, which gives $x=y+1$, and substituting that value of $x$ into the second equation which gives $y+1+y=(y+1)^{2}-3$. Rewriting this equation by adding like terms and expanding $ (y+1)^{2} $ gives $2y+1=y^{2}+2y-2$. Subtracting $2y$ from both sides of this equation gives $ 1=y^{2}-2$. Adding to $2$ to both sides of this equation gives $ 3=y^{2}$. Therefore, it follows that $ y=\\pm\\sqrt{3}$. Substituting $ \\sqrt{3}$ for $y$ in the first equation yields $ x-\\sqrt{3}=1$. Adding $ \\sqrt{3}$ to both sides of this equation yields $ x=1+\\sqrt{3}$. Therefore, the ordered pair $ (1+\\sqrt{3},\\sqrt{3}) $ is a solution to the given system of equations.

Choice B is incorrect. Substituting $ \\sqrt{3}$ for x and $ -\\sqrt{3}$ for y in the first equation yields $ \\sqrt{3}-(-\\sqrt{3})=1$, or $2\\sqrt{3}=1$, which isn't a true statement. Choice C is incorrect. Substituting $ 1+\\sqrt{5} $ for x and $ \\sqrt{5}$ for y in the second equation yields $ (1+\\sqrt{5})+\\sqrt{5}=(1+\\sqrt{5})^{2}-3$, or $1+2\\sqrt{5}=2\\sqrt{5}+3$, which isn't a true statement. Choice D is incorrect. Substituting $ \\sqrt{5}$ for x and $ (-1+\\sqrt{5}) $ for y in the second equation yields $ \\sqrt{5}+(-1+\\sqrt{5})=(\\sqrt{5})^{2}-3$, or $2\\sqrt{5}-1=2$, which isn't a true statement.

"},"encodingFormat":"text/markdown","text":"

$x-y=1$

$x+y=x^{2}-3$

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

","suggestedAnswer":[{"@type":"Answer","position":1,"encodingFormat":"text/html","text":"

$(\\sqrt{3},-\\sqrt{3})$

"},{"@type":"Answer","position":2,"encodingFormat":"text/html","text":"

$(1+\\sqrt{5},\\sqrt{5})$

"},{"@type":"Answer","position":3,"encodingFormat":"text/html","text":"

$(\\sqrt{5},-1+\\sqrt{5})$

"}],"acceptedAnswer":{"@type":"Answer","position":0,"encodingFormat":"text/html","text":"

$(1+\\sqrt{3},\\sqrt{3})$

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

"},"answerExplanation":{"@type":"Comment","text":"

"}}},"url":"https://classnotes.xyz/mcqs/sat-math-advanced-math-multiple-choice-question-10"}

$x-y=1$$x+y=x^{2}-3$Which ordered pair is a solution to the system of equations above?

$x-y=1$$x+y=x^{2}-3$Which ordered pair is a solution to the system of equations above?

$x-y=1$$x+y=x^{2}-3$Which ordered pair is a solution to the system of equations above?

$x-y=1$

$x+y=x^{2}-3$

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

$x-y=1$

$x+y=x^{2}-3$

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

$x-y=1$

$x+y=x^{2}-3$

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