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

$y=x^{2}+2x+1$

$x+y+1=0$

If $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are the two solutions to the system of equations above, what is the value of $y_{1}+y_{2}$?

","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-86"}],"about":{"@type":"Thing","name":"Math, Advanced Math, Duplicate"},"educationalLevel":"intermediate","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","assesses":["Math","Advanced Math","Duplicate"],"name":"

$y=x^{2}+2x+1$

$x+y+1=0$

If $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are the two solutions to the system of equations above, what is the value of $y_{1}+y_{2}$?

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

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 $-x-1=x^{2}+2x+1$. Adding $x + 1$ to both sides of the equation yields $x^{2}+3x+2=0$. 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, $(x_{1},y_{1})=(-2,1)$, $(x_{2},y_{2})=(-1,0)$, and $y_{1} + y_{2} = 1 + 0 = 1$.

Choice A is incorrect. The solutions to the system of equations are $(x_{1},y_{1})=(-2,1)$ and $(x_{2},y_{2})=(-1,0)$. 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.

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

$y=x^{2}+2x+1$

$x+y+1=0$

If $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are the two solutions to the system of equations above, what is the value of $y_{1}+y_{2}$?

","suggestedAnswer":[{"@type":"Answer","position":0,"encodingFormat":"text/html","text":"

$-3$

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

$-2$

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

\$-1\$

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

$1$

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

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

"}}},"url":"https://classnotes.xyz/mcqs/sat-math-advanced-math-multiple-choice-question-86"}

$y=x^{2}+2x+1$$x+y+1=0$If $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are the two solutions to the system of equations above, what is the value of $y_{1}+y_{2}$?

$y=x^{2}+2x+1$$x+y+1=0$If $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are the two solutions to the system of equations above, what is the value of $y_{1}+y_{2}$?

$y=x^{2}+2x+1$$x+y+1=0$If $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are the two solutions to the system of equations above, what is the value of $y_{1}+y_{2}$?

$y=x^{2}+2x+1$

$x+y+1=0$

If $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are the two solutions to the system of equations above, what is the value of $y_{1}+y_{2}$?

$y=x^{2}+2x+1$

$x+y+1=0$

If $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are the two solutions to the system of equations above, what is the value of $y_{1}+y_{2}$?

$y=x^{2}+2x+1$

$x+y+1=0$

If $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are the two solutions to the system of equations above, what is the value of $y_{1}+y_{2}$?