2y(2y - 1)(2y - 3)
2y(2y - 1)(2y - 3)
2y(3x + 1)
2y^2+5y+2=(2y+1)(y+2)
x = 11 - 2y and 14 - 2y. This means that 11 = 14. Is there a typo in the question?
2y(2y - 1)(2y - 3)
2y(2y - 1)(2y - 3)
2y(3x + 1)
2y + 1
2y^2+5y+2=(2y+1)(y+2)
7(t - 2y)(t + 2y)
x = 11 - 2y and 14 - 2y. This means that 11 = 14. Is there a typo in the question?
-6y + 14 + 4y = 32-2y = 32 - 14-2y = 18y = - 9
The common factor is 2.
2+2y+x+xy=(x+2)(y+1)
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
(2y - 5)(y - 2)