Solve tx^2-2/3x-1/3=08y

$$t x ^ { 2 } - \frac { 2 } { 3 } x - \frac { 1 } { 3 } = 08 y$$

Solve for t

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

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply 0 and 8 to get 0.

$$tx^{2}-\frac{2}{3}x-\frac{1}{3}=0y$$

Anything times zero gives zero.

$$tx^{2}-\frac{2}{3}x-\frac{1}{3}=0$$

Add \frac{2}{3}x to both sides. Anything plus zero gives itself.

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

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

$$tx^{2}=\frac{2}{3}x+\frac{1}{3}$$

The equation is in standard form.

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

Divide both sides by x^{2}.

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

Dividing by x^{2} undoes the multiplication by x^{2}.

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{3t+1}+1}{3t}\text{; }x=\frac{-\sqrt{3t+1}+1}{3t}\text{, }&t\neq 0\\x=-\frac{1}{2}\text{, }&t=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{3t+1}+1}{3t}\text{; }x=\frac{-\sqrt{3t+1}+1}{3t}\text{, }&t\neq 0\text{ and }t\geq -\frac{1}{3}\\x=-\frac{1}{2}\text{, }&t=0\end{matrix}\right.$$