Solve for b
\left\{\begin{matrix}b=\frac{2z}{3x^{2}}\text{, }&x\neq 0\\b\in \mathrm{R}\text{, }&z=0\text{ and }x=0\end{matrix}\right.
Solve for x
\left\{\begin{matrix}x=\frac{\sqrt{\frac{6z}{b}}}{3}\text{; }x=-\frac{\sqrt{\frac{6z}{b}}}{3}\text{, }&\left(z\geq 0\text{ and }b>0\right)\text{ or }\left(z\leq 0\text{ and }b<0\right)\\x\in \mathrm{R}\text{, }&z=0\text{ and }b=0\end{matrix}\right.
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b\times 3x^{2}=2z
Add 2z to both sides. Anything plus zero gives itself.
3x^{2}b=2z
The equation is in standard form.
\frac{3x^{2}b}{3x^{2}}=\frac{2z}{3x^{2}}
Divide both sides by 3x^{2}.
b=\frac{2z}{3x^{2}}
Dividing by 3x^{2} undoes the multiplication by 3x^{2}.
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