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20 August, 18:23

What is the product of the expressions? Assume y ≠ 0. (2y-1/y2) (3y2/7)

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  1. 20 August, 18:39
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    (2y-1/y_2) (3y_2/7)

    = 2y * (3y_2/7) - (1/y_2) (3y_2/7)

    = 6y*y_2/7 - 3/7 = (6y*y_2 - 3) / 7

    Note: I'm not sure if the 2y-1 was written correctly. If it were intended as the entire thing being a numerator or as 2y_1, this answer is inaccurate.
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