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If (6y − 11)(6y 11) = ay^2 − b, what is the value of a? the value of a is ____________.

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(6y - 11)(6y + 11) = ay² - b |use (a - b)(a + b) = a² - b²

(6y)² - 11² =ay² - b

36y² - 121 = ay² - b |add b to both sides

ay² = 36y² - 121 + b |divide both sides by y² ≠ 0

a = (36y² - 121 + b)/y²


\boxed{a=(36y^2-121+b)/(y^2)}
User Buneme Kyakilika
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