y(y-2) i am not sure i was stuck ... that is what i have put :S because if you multiply is out again you have yxy= y^2 and yx2 +2y
5
if you take a factor of (Y) you end up with Y(X+5Z-2).
2(x+y) To find the answer above, look at the question (2x + 2y), and see what the common factors are. In this case, it is "2", which is present in both "x" and "y". Put the "2" on the outside, and write the rest of the equation.
(2y + 1) + (y^2 + 1) = 2y + 1 + y^2 + 1 = y^2 + 2y + 2
2y
x2 - 4y2 = 16∴ (x - 2y)(x + 2y) = 162y - x = 2∴ x = 2y - 2∴ ([2y - 2] - 2y)([2y - 2] + 2y) = 16∴ (y - 1 - y)(y - 1 + y) = 16∴ -1(2y - 1) = 16∴ 1 - 2y = 16∴ -2y = 15∴ y = -7.52y - x = -2∴ -15 - x = -2∴ x = -13So the point of intersection is (-13, -7.5)
2x2 + 2y - 8 = 0 2(x2 + y - 4) = 0 and other than removing the two outside the bracket, this cannot be factorised further, nor simplified.
(4x)(2y) would be factored as such: 4x = 2 • 2 • x 2y = 2 • y 2(2x)(y)
3xy-2y=0 3xy=2y y=2y (3x) y/2y=3x 1/2=3x multiply across by 2 1=6x 1/6=x therefore substituting x=1/6 into 3xy-2y; 3(1/6)y-2y=0 1/2y=2y y=2y/0.5 0.5 aka 1/2 y=1
2y.2y
(A) 3x + 2y = -2 (B) 6x - y = 6 (A) + 2(B): 3x + 2y + 12x - 2y = -2 + 12 or 15x = 10 or x = 2/3 substitute this value of x in (A) 2 + 2y = -2 2y = -4 and y = -2 Answer: x = 2/3, y = -2
Y squared (Y^2)