what are the solutions to the following system?

$\begin{cases} -2x^2 + y = -5 \\ y = -3x^2 + 5 \end{cases}$

(0, 2)
(1, -2)
$(\sqrt{2}, -1)$ and $(-\sqrt{2}, -1)$
$(\sqrt{5}, -10)$ and $(-\sqrt{5}, -10)$

Question

what are the solutions to the following system?

$\begin{cases} -2x^2 + y = -5 \\ y = -3x^2 + 5 \end{cases}$

(0, 2)
(1, -2)
$(\sqrt{2}, -1)$ and $(-\sqrt{2}, -1)$
$(\sqrt{5}, -10)$ and $(-\sqrt{5}, -10)$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Systems of Equations", "Solving Quadratic Equations" ], "new_concepts": [ "Intersection of Quadratic Graphs" ], "current_concepts": [ "Systems of Equations", "Solving Quadratic Equations", "Intersection of Quadratic Graphs" ] } ## Express y from the first equation $$ \begin{aligned} -2x^2 + y &= -5 \ y &= 2x^2 - 5 \end{aligned} $$ ## Equate the two expressions for y $$ \begin{aligned} 2x^2 - 5 &= -3x^2 + 5 \ 5x^2 &= 10 \ x^2 &= 2 \ x &= \pm\sqrt{2} \end{aligned} $$ ## Solve for the corresponding y-values $$ \begin{aligned} y &= 2(\pm\sqrt{2})^2 - 5 \ y &= 2(2) - 5 \ y &= -1 \end{aligned} $$

(A) $(0, 2)$

(B) $(1, -2)$

(C) $(\sqrt{2}, -1)$ and $(-\sqrt{2}, -1)$

(D) $(\sqrt{5}, -10)$ and $(-\sqrt{5}, -10)$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Intersection of Quadratic Graphs" ] }

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QUESTION IMAGE

Question

Explanation:

Express y from the first equation

Equate the two expressions for y

Solve for the corresponding y-values

Answer: