multiply by 2y/x
Y = - (1/3)2y = - 1/9
if you take a factor of (Y) you end up with Y(X+5Z-2).
(2y - 5)(y - 2)
(2y+1)(y−5)
Y squared (Y^2)
2y squared * y cubed= 2y^5 (2y to the fifth power)
0
9y cubed plus 2y squared
(y - 8)(y - 2)
multiply by 2y/x
1st of all, if you mean 9-y2 its simply a difference between two squares formula (3-y) (3+y)
(2y + 1) + (y^2 + 1) = 2y + 1 + y^2 + 1 = y^2 + 2y + 2
Y = - (1/3)2y = - 1/9
-6
if you take a factor of (Y) you end up with Y(X+5Z-2).
If: x = 2y-2 and x^2 = y^2+7 Then: (2y-2)^2 = y^2+7 So: (2y-2)(2y-2) = y^2+7 It follows: 4y^2-y^2-8y+4-7 = 0 => 3y^2-8y-3 = 0 Solving the above quadratic equation: y = -1/3 or y = 3 Solutions: when y = 3 then x = 4 and when y = -1/3 then x = -8/3