I think you mean (3x ± 1)^2 = (3x+1)(3x-1)?
In the equation 3x plus 7 equals 13, x is equal to 2.
X= -2 ! Since the equation must equal zero - 17+3x must equal 11. The variable 3x HAS to equal -6... therefore x is equal to minus 2.
3x = x + 4 x = 2 3x2=6 2+4=6
3x + 2 = 3x + 6 This is not possible.
I think you mean (3x ± 1)^2 = (3x+1)(3x-1)?
In the equation 3x plus 7 equals 13, x is equal to 2.
X= -2 ! Since the equation must equal zero - 17+3x must equal 11. The variable 3x HAS to equal -6... therefore x is equal to minus 2.
It is equal to: x^2 +3x -10
3x = x + 4 x = 2 3x2=6 2+4=6
3x + 2 = 3x + 6 This is not possible.
y = -1x/2 3x + 6y = 0 -3x -3x __________ 6y = 0 - 3x --- -------- 6 6 y= -3x/6 y= -1x/2
(x2+ 3x + 2) / (x + 2) = (x + 1)(x + 2) / (x + 2) = x + 1; but with the following important proviso: (x + 2) can not equal zero; otherwise we are dividing by zero. Thus, (x2+ 3x + 2) / (x + 2) = x + 1, if x ≠ -2; and (x2+ 3x + 2) / (x + 2) is undefined, if x = -2.
3x=2 Divide the x over and you get 3 divided by 2 x=3/2
3x+2.
x^2 + 3x + 2/(-3x) - x^2 - 2=3x-2+2/(-3x)=3x-2-2/(3x)
3x-3x+4x^2+8x=4x^2+8x