Solve for v (complex solution)
\left\{\begin{matrix}v=-\frac{y^{t}-2y}{2t}\text{, }&t\neq 0\\v\in \mathrm{C}\text{, }&\left(y=\frac{1}{2}\text{ or }y=0\right)\text{ and }t=0\end{matrix}\right.
Solve for v
\left\{\begin{matrix}v=-\frac{y^{t}-2y}{2t}\text{, }&\left(t\neq 0\text{ and }y>0\right)\text{ or }\left(y=0\text{ and }t>0\right)\text{ or }\left(t\neq 0\text{ and }y<0\text{ and }Denominator(t)\text{bmod}2=1\right)\\v\in \mathrm{R}\text{, }&y=\frac{1}{2}\text{ and }t=0\end{matrix}\right.
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vt+\frac{1}{2}y^{t}=y
Swap sides so that all variable terms are on the left hand side.
vt=y-\frac{1}{2}y^{t}
Subtract \frac{1}{2}y^{t} from both sides.
tv=-\frac{y^{t}}{2}+y
The equation is in standard form.
\frac{tv}{t}=\frac{-\frac{y^{t}}{2}+y}{t}
Divide both sides by t.
v=\frac{-\frac{y^{t}}{2}+y}{t}
Dividing by t undoes the multiplication by t.
vt+\frac{1}{2}y^{t}=y
Swap sides so that all variable terms are on the left hand side.
vt=y-\frac{1}{2}y^{t}
Subtract \frac{1}{2}y^{t} from both sides.
tv=-\frac{y^{t}}{2}+y
The equation is in standard form.
\frac{tv}{t}=\frac{-\frac{y^{t}}{2}+y}{t}
Divide both sides by t.
v=\frac{-\frac{y^{t}}{2}+y}{t}
Dividing by t undoes the multiplication by t.
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