Solve y=1/3x^3-1/2ax^2+x-1

$$y = \frac{ 1 }{ 3 } { x }^{ 3 } - \frac{ 1 }{ 2 } a { x }^{ 2 } +x-1$$

Solve for a

$$\left\{\begin{matrix}a=\frac{2\left(x^{3}+3x-3y-3\right)}{3x^{2}}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&y=-1\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{3}x^{3}-\frac{1}{2}ax^{2}+x-1=y$$

Subtract x from both sides.

$$\frac{1}{3}x^{3}-\frac{1}{2}ax^{2}-1=y-x$$

Add 1 to both sides.

$$\frac{1}{3}x^{3}-\frac{1}{2}ax^{2}=y-x+1$$

Subtract \frac{1}{3}x^{3} from both sides.

$$-\frac{1}{2}ax^{2}=y-x+1-\frac{1}{3}x^{3}$$

The equation is in standard form.

$$\left(-\frac{x^{2}}{2}\right)a=-\frac{x^{3}}{3}+y-x+1$$

Divide both sides by -\frac{1}{2}x^{2}.

$$\frac{\left(-\frac{x^{2}}{2}\right)a}{-\frac{x^{2}}{2}}=\frac{-\frac{x^{3}}{3}+y-x+1}{-\frac{x^{2}}{2}}$$

Dividing by -\frac{1}{2}x^{2} undoes the multiplication by -\frac{1}{2}x^{2}.

$$a=\frac{-\frac{x^{3}}{3}+y-x+1}{-\frac{x^{2}}{2}}$$

Divide y-x-\frac{x^{3}}{3}+1 by -\frac{1}{2}x^{2}.

$$a=-\frac{2\left(3+3y-3x-x^{3}\right)}{3x^{2}}$$