I think you mean (3x ± 1)^2 = (3x+1)(3x-1)?
x=1, y=5 is a legitimate answer.
YES ! When x = -1 then y = 3(-1) + 2 = -3 + 2 = -1
2x+3 = 3x+5 2x-3x = 5-3 -x = 2 x = -2
x + 2 = 3x-1 -2x = -3x = 1,5
If you mean y = 3x+2 then (-1, -1) is a solution of the given straight line equation
The solution to 5 = 3x + 2 is found by first subtracting 2:3 = 3x.Then dividing by 3:x = 1.
I think you mean (3x ± 1)^2 = (3x+1)(3x-1)?
What point is a solution to y>3x-2
x=1, y=5 is a legitimate answer.
YES ! When x = -1 then y = 3(-1) + 2 = -3 + 2 = -1
If you mean: y = 3x+2 then (-1, -1) will satisfy the straight line equation
If that's -9x^3 - 3x^2 + 3x + 1, that factors to (1 - 3x^2)(1 + 3x)
X= -1 1*2 0-1 -1-2 -3/3 -1
2x+3 = 3x+5 2x-3x = 5-3 -x = 2 x = -2
9x2 + 6x + 1 = 9x2 + 3x + 3x + 1 = 3x(3x + 1) + 1(3x + 1) = (3x + 1)(3x + 1) or (3x +1)2
The mathematical symbols did not appear so I am assuming that your question is if the point (-1,-1) is a solution of y = 3x + 2, in that case, substitute -1 for y and -1 for x and you get -1 = 3(-1) + 2 or -1 = -3 + 2 or -1 = -1 Thus if I assumed the question correctly, then -1,-1 is a solution to y = 3x + 2