If you mean to say (x^2)-2x+4=0 then there is no solution for x because the equation never makes it to 0 If you mean to say 2x-2x+4=0 then there is no solution for x because it is a horizontal line at Y=4
If you have -2x = 0, to solve for x, divide each side by two. That results in x = -0/2, or x = 0. So there is only one solution.
y = 0x = y + 2plug 0 in for yx = 0 + 2x = 2
1 +/- the square root of 3
x = 0 & y = -3
Yes it is!
x = 0
The two equations are linear.
If you mean to say (x^2)-2x+4=0 then there is no solution for x because the equation never makes it to 0 If you mean to say 2x-2x+4=0 then there is no solution for x because it is a horizontal line at Y=4
Yes.
If you have -2x = 0, to solve for x, divide each side by two. That results in x = -0/2, or x = 0. So there is only one solution.
-2
y = 0x = y + 2plug 0 in for yx = 0 + 2x = 2
The solution is:- x = 0 and y = 5
1 +/- the square root of 3
7 and (-3/2)
4x2-4x-3 = 0 (2x-3)(2x+1) = 0 x = 3/2 or x = -1/2