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2+2y+x+xy=(x+2)(y+1)
I understand you want [(2/x) + (1/y)] as a single fraction. The common denominator is (xy): 2/x = (2y)/(xy); and 1/y = x/(xy), so the answer is (2y + x) / (xy)
if you take a factor of (Y) you end up with Y(X+5Z-2).
2+2y+x+xy
(xy - 7)(x^2y^2 + 7xy + 49)
2+2y+x+xy=(x+2)(y+1)
(x + 2)(y - 3)
xy(x - 2y)(x + 2y)
0
(y - 3)(x + 2y)
I understand you want [(2/x) + (1/y)] as a single fraction. The common denominator is (xy): 2/x = (2y)/(xy); and 1/y = x/(xy), so the answer is (2y + x) / (xy)
if you take a factor of (Y) you end up with Y(X+5Z-2).
2+2y+x+xy
(1 - x4y4) = (1 + x2y2)*(1 - x2y2) = (1 + x2y2)*(1 + xy)*(1 - xy)
(xy - 7)(x^2y^2 + 7xy + 49)
(xy + 7)(x^2y^2 - 7xy + 49)
2 plus 2y plus x plus x y = 2 plus3y plus 2x (find how many X's there are and add them together and ten Y's and then single numbers)