Solve t(-1/6t^2+1/2t)x-1/3=1/3t^2-t

$$t ( - \frac { 1 } { 6 } t ^ { 2 } + \frac { 1 } { 2 } t ) x - \frac { 1 } { 3 } = \frac { 1 } { 3 } t ^ { 2 } - t$$

Solve for x

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

$$t\neq 3\text{ and }t\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply t by -\frac{1}{6}t^{2}+\frac{1}{2}t.

$$\left(-\frac{1}{6}t^{3}+\frac{1}{2}t^{2}\right)x-\frac{1}{3}=\frac{1}{3}t^{2}-t$$

Use the distributive property to multiply -\frac{1}{6}t^{3}+\frac{1}{2}t^{2} by x.

$$-\frac{1}{6}t^{3}x+\frac{1}{2}t^{2}x-\frac{1}{3}=\frac{1}{3}t^{2}-t$$

Add \frac{1}{3} to both sides.

$$-\frac{1}{6}t^{3}x+\frac{1}{2}t^{2}x=\frac{1}{3}t^{2}-t+\frac{1}{3}$$

Combine all terms containing x.

$$\left(-\frac{1}{6}t^{3}+\frac{1}{2}t^{2}\right)x=\frac{1}{3}t^{2}-t+\frac{1}{3}$$

The equation is in standard form.

$$\left(-\frac{t^{3}}{6}+\frac{t^{2}}{2}\right)x=\frac{t^{2}}{3}-t+\frac{1}{3}$$

Divide both sides by -\frac{1}{6}t^{3}+\frac{1}{2}t^{2}.

$$\frac{\left(-\frac{t^{3}}{6}+\frac{t^{2}}{2}\right)x}{-\frac{t^{3}}{6}+\frac{t^{2}}{2}}=\frac{\frac{t^{2}}{3}-t+\frac{1}{3}}{-\frac{t^{3}}{6}+\frac{t^{2}}{2}}$$

Dividing by -\frac{1}{6}t^{3}+\frac{1}{2}t^{2} undoes the multiplication by -\frac{1}{6}t^{3}+\frac{1}{2}t^{2}.

$$x=\frac{\frac{t^{2}}{3}-t+\frac{1}{3}}{-\frac{t^{3}}{6}+\frac{t^{2}}{2}}$$

Divide \frac{t^{2}}{3}-t+\frac{1}{3} by -\frac{1}{6}t^{3}+\frac{1}{2}t^{2}.

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