2/y + 1 + 3 + 5y
2y^2+5y+2=(2y+1)(y+2)
The multiplicative inverse of 5y -xy + 1 is 1/5y -xy + 1 The additive inverse of 5y - xy + 1 is -5y + xy - 1
(4y - 1)(5y + 2)
(2, 1)
2/y + 1 + 3 + 5y
2y^2+5y+2=(2y+1)(y+2)
5y + 7 = 25y = -5y = -1
The multiplicative inverse of 5y -xy + 1 is 1/5y -xy + 1 The additive inverse of 5y - xy + 1 is -5y + xy - 1
(4y - 1)(5y + 2)
x + 5y = 10 5y = -x + 10 y = -x/5 + 2 m = -1/5 or .2
5y-2y= 3y+2 5y-2y-3y= 2 0y= 2
(2, 1)
(1, -1)
3x + 5y = -1 2x - 5y = 16 Add the two equations: 5x = 15 or x = 3 Substitute for x in the first equation: 3*3 + 5y = -1 15 + 5y = -1 5y = -10 so y = -10/5 = -2 The solution is (x, y) = (3, -2)
(-1, 2)
5y+8=7y subtract 5y from each side , 5y+8=7y -5y -5y 8=2y /2 /2 4=y