\[ 2a + b = 6 \]

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• \( h(2) = 15 \) gives us \( a(2)^2 + b(2) + 3 = 15 \) which simplifies to \( 4a + 2b + 3 = 15 \). Thus, \( 4a + 2b = 12 \).
• \( a + b = 5 \)
• \( 4a + 2b = 12 \)
• \[ (2a + b) - (a + b) = 6 - 5 \]
• \[ 1 + b = 5 \]
• \[ b = 4 \]

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