\2x + 3y = 12\quad x - y = 1\

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{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Systems of Equations", "Elimination Method" ], "new_concepts": [], "current_concepts": [ "Systems of Equations", "Elimination Method", "Substitution Method" ] } ## Set up the system of equations $$ \begin{cases} 2x + 3y = 12 \quad \text{--- (1)} \ x - y = 1 \quad \text{--- (2)} \end{cases} $$ ## Eliminate the variable y Multiply equation (2) by 3 and add it to equation (1): $$ \begin{aligned} 3(x - y) &= 3(1) \ 3x - 3y &= 3 \quad \text{--- (3)} \end{aligned} $$ $$ \begin{aligned} (2x + 3y) + (3x - 3y) &= 12 + 3 \ 5x &= 15 \ x &= 3 \end{aligned} $$ ## Solve for y Substitute $x = 3$ into equation (2): $$ \begin{aligned} 3 - y &= 1 \ y &= 2 \end{aligned} $$ $x = 3$, $y = 2$ { "subject": "Mathematics", "question_type": "Free Response", "knowledge_point": [ "Mathematics", "Algebra", "Systems of Equations" ] }

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Explanation:

Set up the system of equations

Eliminate the variable y

Solve for y

Answer: