2y squared * y cubed= 2y^5 (2y to the fifth power)
If: 3x+2y = 5x+2y = 14 Then: 3x+2y = 14 and 5x+2y =14 Subtract the 1st equation from the 2nd equation: 2x = 0 Therefore by substitution the solutions are: x = 0 and y = 7
7+4xz+52xz+2y=7+56xz+2y
2 (2x2 + 2x + xy + y ) it's 2 (2x + y)(x + 1) if you're doing A+
12x-4y+9z+2y =12x-2y+9z
2y squared * y cubed= 2y^5 (2y to the fifth power)
-6
2x^2y^3(4x^2 - 3xy + 1)
2y + 1
2y^2+5y+2=(2y+1)(y+2)
y^(3) - 2y^(2) + 4y Factor out 'y' Hence y(y^2 - 2y + 4) This will not factor any further.
6xyz(3x + 2y + z)
(2y - 5)(y - 2)
(2y + 1) + (y^2 + 1) = 2y + 1 + y^2 + 1 = y^2 + 2y + 2
(2y)3 = 8y3
If that's -18y2, the answer is (3x - 2y)(x + 9y)
3y squared - 2y = 1