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(A): 4x + y = 7 (B): 2x - 3y = -7 2*(B) 4x - 6y = -14 Subtract from (A) 7y = 21 Divide both sides by 7: y = 3 Substitute in (A): 4x + 3 = 7 ie 4x = 4 or x = 1 Answer: x = 1 , y = 3
12x2 + 7x + 1 = 12x2 + 3x + 4x + 1 = 3x*(4x + 1) + 1*(4x + 1) = (4x + 1)*(3x + 1)
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You would probably use a power-reduction trig identity to solve this equation. This states that sin2(x) = (1 - cos(2x))/2 Therefore, sin2(2x) = (1 - cos(4x))/2, or (1/2)(1 - cos(4x)) So, ∫ (1/2)(1 - cos(4x)) dx = (1/2) ∫ (1 - cos(4x)) dx. Then, ∫ (1-cos4x)dx = x - (1/4)sin(4x) + c Now, multiply that by (1/2) to get: (x - (sin(4x)/4) + c)/2 Since c is an arbitrary constant, we have: ½(x - sin(4x) / 4) + c OR 1/8 * (4x - sin(4x)) + c
-1-2x-4x=19-6x-1 = -19x = 3