x+xy=8 xy=-x+8 y=-1+8/x
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)
(1/x) - (1/y) = (1/z) Get the left-hand side over a common denominator:- (y-x)/xy = 1/z Take the reciprocal of both sides:- z = xy / (y-x)
The multiplicative inverse of 5y -xy + 1 is 1/5y -xy + 1 The additive inverse of 5y - xy + 1 is -5y + xy - 1
x+xy=8 xy=-x+8 y=-1+8/x
xy + x + y + 1 = (x + 1)(y + 1).
(x-y) + (xy - 1) = (x - 1)(y + 1)
y=(x+2)/(x+10) xy+10y=x+2 xy-x=2-10y x(y-1)=2-10y x=(2-10y)/(y-1)
2+2y+x+xy=(x+2)(y+1)
4
0
To solve the given terms is almost impossible in the absence of any plus, minus and equality signs.
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)
(1/x) - (1/y) = (1/z) Get the left-hand side over a common denominator:- (y-x)/xy = 1/z Take the reciprocal of both sides:- z = xy / (y-x)
The multiplicative inverse of 5y -xy + 1 is 1/5y -xy + 1 The additive inverse of 5y - xy + 1 is -5y + xy - 1
If that's xy + x^2y, it factors to xy(x + 1) If it's something else, please re-submit your question with any plus signs written out.