4ac- 62 { Acos (VABC-B³² x) 2a y(x) = Ã cos + B sin y(x) = A cos bx ²x)} exp (-26) 2a 4ac- 62 2a We could also, for suitable new constants A and , write x +0) exp(-26) √4ac- 6² 2a

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A:

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Q: 7) -6.x+ 6y = 6 -6x + 3y = -12

A: Using Substitution,

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A: What should you do to get the two terms with variables on the left side of the equation?

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A: Approach to solving the question: Detailed explanation:

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A: To solve the given equation.

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Q: (5и — 1)(Зи? +2) du

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A:

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