what are the solutions of the following system?

$\begin{cases} 10x^2 - y = 48 \\ 2y = 16x^2 + 48 \end{cases}$

$(2\sqrt{3}, 120)$ and $(-2\sqrt{3}, 120)$
$(2\sqrt{3}, 120)$ and $(-2\sqrt{3}, -72)$
$(6, 312)$ and $(-6, 312)$
$(6, 312)$ and $(-6, -264)$

Question

what are the solutions of the following system?

$\begin{cases} 10x^2 - y = 48 \\ 2y = 16x^2 + 48 \end{cases}$

$(2\sqrt{3}, 120)$ and $(-2\sqrt{3}, 120)$
$(2\sqrt{3}, 120)$ and $(-2\sqrt{3}, -72)$
$(6, 312)$ and $(-6, 312)$
$(6, 312)$ and $(-6, -264)$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [], "current_concepts": [ "Systems of Equations", "Solving Quadratic Equations" ] } ## Express y in terms of x^2 $$ \begin{aligned} 10x^2 - y &= 48 \implies y = 10x^2 - 48 \end{aligned} $$ ## Substitute y into the second equation $$ \begin{aligned} 2(10x^2 - 48) &= 16x^2 + 48 \ 20x^2 - 96 &= 16x^2 + 48 \ 4x^2 &= 144 \ x^2 &= 36 \ x &= \pm 6 \end{aligned} $$ ## Solve for y using the values of x $$ \begin{aligned} y &= 10(6)^2 - 48 = 10(36) - 48 = 360 - 48 = 312 \ y &= 10(-6)^2 - 48 = 10(36) - 48 = 360 - 48 = 312 \end{aligned} $$

(A) $(2\sqrt{3}, 120)$ and $(-2\sqrt{3}, 120)$

(B) $(2\sqrt{3}, 120)$ and $(-2\sqrt{3}, -72)$

(C) $(6, 312)$ and $(-6, 312)$

(D) $(6, 312)$ and $(-6, -264)$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Systems of Equations" ] }

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QUESTION IMAGE

Question

Explanation:

Express y in terms of x^2

Substitute y into the second equation

Solve for y using the values of x

Answer: