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, because if you sub in for x and y, you get 4-0=4 and this is true
-2x - 2 = 4-2x = 4 + 2-2x = 6x = 6/-2x = -3Check:-2(-3) - 2 = 46 - 2 = 44 = 4
x = 0
If you mean: -2x = -4 then the value of x is 2
If you mean: 2x-y = -4 then they are at (-2, 0) and (0, 4)
If: y = x^2 -2x +4 and y = 2x^2 -4x +4 Then: 2x^2 -4x +4 = x^2 -2x +4 Transposing terms: x^2 -2x = 0 Factorizing: (x-2)(x+0) => x = 2 or x = 0 Therefore by substitution points of intersect are at: (2, 4) and (0, 4)
Divide both sides by two. x = 2 2x=4 2x/2=4/2 2x/2=1x=x 4/2=2 x=2
y=2x-4 y=2x-5 y=1 1=2x-4 -2x = -5 x=2/5 the solution is (x,y) = (2/5,1)
There is no solution to: 4(2x + 2) + 12 > 1004(2x + 2) + 12 > 100 For 4(2x + 2) + 12 > 1004(2x + 2) + 12 to have a solution x < -1 which makes 1004(2x + 2) + 12 < 12, BUT 1004(2x + 2) + 12 is supposed to be greater than 100. Perhaps there is a missing operator in the digits of 1004?
2x^2 + 3x - 20 = 0 is 2x^2 + 8x - 5x -20 = 0 is 2x(x + 4) -5(x + 4) = 0 is (2x - 5)(x +4) = 0 Hence 2x -5 = 0 which gives x = 2.5 or x + 4 = 0 which gives x = -4
If you mean: 4(2x-2) = 4-4(x-3) then the value of x works out as 2