Since x and y are variables, the sum of 5x and 5y could be literally any number.
a(x5) + b(x5) + c(x5) + d(x5) + e(x5) = abcde(a+b+c+d+e) x5 = abcdeThis equation has at least 5 variables. To solve for all of them requires at least 4 more equations.
It is 3(2y+3) when factorized
It is: 2(11x+3) when factorized
x2+bx+ax+ab = x2+ax+bx+ab = x(x+a)+b(x+a) = (x+a)(x+b)
Since x and y are variables, the sum of 5x and 5y could be literally any number.
4x-y2=2xy 2x ? y5 if its plus its 7 xy
Factors of x5 . . . 1, x, x2, x3, x4, and x5, as well as any factors of 'x', and their powers. Factors of y5 . . . 1, y, y2, y3, y4, and y5, as well as any factors of 'y', and their powers.
a(x5) + b(x5) + c(x5) + d(x5) + e(x5) = abcde(a+b+c+d+e) x5 = abcdeThis equation has at least 5 variables. To solve for all of them requires at least 4 more equations.
k(7k + 9)
y5(x3 - 8)
-2(x + 5)(x - 6)
It is 3(2y+3) when factorized
It is: 2(11x+3) when factorized
It is: 12(2x+3) when factored
6-y5 = 1
The additive inverse is x5 + 2x - 2.