So, we have 3x = 4 for some number x. We will solve for x. 3x = 4 (1/3)*(3x) = (1/3)*(4) [multiplied by (1/3) on both sides] x = 4/3
3x - x = (3 - 1)x = 2x
{3x +y =1 {x+y= -3
3X^2 + 9X/3X factor out 3X 3X(X + 3)/3X 3X/3X = 1, so... X + 3 is the simplest form here
3x+3=0 -3 = -3 = 3x=-3 /3 /3 x= -1
x-2=3x+1 x=3x+1+2 x-3x=3 -2x=3 x=-3/2
y = (3x + 1)⁵(3 - x)³ The stationary points are where dy/dx = 0 Note the use of: If u and v are functions of x: d/dx (uv) = v du/dx + u dv/dx d/dx uⁿ = n uⁿ⁻¹ du/dx dy/dx = (3 - x)³ d/dx (3x + 1)⁵ + (3x + 1)⁵ d/dx (3 - x)³ = (3 - x)³ 5 (3x + 1)⁴ d/dx (3x + 1) + (3x + 1)⁵ 3 (3 - x)² d/dx (3 - x) = 5 (3 - x)³ (3x + 1)⁴ 3 + 3 (3x + 1)⁵ (3 - x)² (-1) = 3 (3x + 1)⁴ (3 - x)² (5 (3 - x) - (3x + 1)) = 3 (3x + 1)⁴ (3 - x)² (15 - 5x - 3x - 1) = 3 (3x + 1)⁴ (3 - x)² (14 - 8x) = 6 (3x + 1)⁴ (3 - x)² (7 - 4x) → 3x + 1 = 0 → x = -1/3 or 3 - x = 0 → x = 3 or 7 - 4x = 0 → x = 7/4
(x^3 + 2x^2 + 3x - 6)/(x - 1) add and subtract x^2, and write -6 as (- 3) + (-3) = (x^3 - x^2 + x^2 + 2x^2 - 3 + 3x - 3)/(x - 1) = [(x^3 - x^2) + (3x^2 - 3) + (3x - 3)]/(x - 1) = [x^2(x - 1) + 3(x^2 - 1) + 3(x - 1)]/(x - 1) = [x^2(x - 1) + 3(x - 1)(x + 1) + 3(x - 1)]/(x - 1) = [(x - 1)(x^2 + 3x + 3 + 3)]/(x - 1) = x^2 + 3x + 6
9x2 + 3x / 3x equals (3x + 1).9x2 + 3x divided by x is 9x + 3 , and then divided by 3 yields the 3x +1.
(3x + 1) = -3 (x + 2x -1) (3x + 1) = -3 (3x - 1) 3x + 1 = -9x + 3 12x = 2 6x = 1 x = 0.16666666666666666666666666666666666666666666666666666666666666666666
3x-1=11 (3x-1)+1=11+1 3x=12 (3x)x(1/3)=12 x (1/3) x=4 x^2+4 16+4 =20
3x2 + x = x*(3x + 1)
4
X2+3x+x+3=x(x+1)+3(x+1)=(x+3)(x+1)
If you mean: (3x-1)(x+1) = 0, then it is 3x squared+2x-1 = 0. Its solutions are: x = 1/3 or x = -1.
So, we have 3x = 4 for some number x. We will solve for x. 3x = 4 (1/3)*(3x) = (1/3)*(4) [multiplied by (1/3) on both sides] x = 4/3
3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3