(8/y-3 - 7/y) - (5/y^2-3y + 2/y)
Pleas explain to me how to do this problem.. I'm so confused on how to even start.
Pleas explain to me how to do this problem.. I'm so confused on how to even start.
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Start by doing your parentheses better. I think your problem is
[8/(y-3) - 7/y] - [5/(y^2-3y) + 2/y]
The common denominator is y(y-3) = y^2 - 3y. Put everything over that, and start getting rid of parentheses:
[8y - 7(y-3) - 5 - 2(y-3)] / (y^2 - 3y) = (8y - 7y + 21 - 5 - 2y + 6) / (y^2 - 3y)
= (-y + 22) / (y^2 - 3y) = (22 - y) / (y^2 - 3y) (Answer)
[8/(y-3) - 7/y] - [5/(y^2-3y) + 2/y]
The common denominator is y(y-3) = y^2 - 3y. Put everything over that, and start getting rid of parentheses:
[8y - 7(y-3) - 5 - 2(y-3)] / (y^2 - 3y) = (8y - 7y + 21 - 5 - 2y + 6) / (y^2 - 3y)
= (-y + 22) / (y^2 - 3y) = (22 - y) / (y^2 - 3y) (Answer)