Divide 21x^2y^3 - 14xy^2 + 7y by 7y.?

Question

show work please

Answer 1

Take each part and divide by 7y

21x^2y^3 by 7y ==> 21/7 = 3 and y^3/y = y^2 so 3x^2y^2

14xy^2 by 7y ==> 14/7 = 2 and y^2/y = y so 2xy

7y/7y = 1

Answer: 3x^2y^2 - 2xy +1

Answer 2

So you want to solve

{21(x^2)(y^3) - 14x(y^2) + 7y} / 7y

Factor the top. Notice that 7y is a common term to everything on the top.

[7y (21(x^2)(y^2) - 14xy + 1)] / 7y

Now that we have a product on the top, we can effectively cancel 7y on the top and bottom, leaving us with

21(x^2)(y^2) - 14xy + 1

Answer 3

(21x^2y^3 - 14xy^2 + 7y )/7y
21x^2y^3/7y-14xy^2/7y+7y/7y
3x^2y^2-2xy+1

Answer 4

(21x^2y^3 - 14xy^2 + 7y)/7y =
7y(3x^2y^2 -2xy + 1)/7y =
3x^2y^2 -2xy + 1

Answer 5

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Answer 6

7y(21x²y³ - 14xy² + 7y) =
3x²y² - 2xy + 1

Answer 7

(21x^2y^3 - 14xy^2 + 7y)/7y
=21x^2y^3/7y - 14xy^2/7y + 7y/7y
=3x^2y^2 - 2xy +1

Answer 8

Each term is divisible by 7. So you end up with:
3x^2y^3 - 2xy^2 + 1y

Each term has 'y' in it. And y = y^1. And 1/y^1 = y^-1
So, all you have to do is subtract 1 from the exponent of each y in the 3 terms. You get:
3x^2y^2 - 2xy^1 + y^0
3x^2y^2 - 2xy + 1