Solve for b
\left\{\begin{matrix}b=\frac{8-a-z}{2z}\text{, }&z\neq 0\\b\in \mathrm{R}\text{, }&a=8\text{ and }z=0\end{matrix}\right.
Solve for a
a=8-z-2bz
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8-2bz-z=a
Swap sides so that all variable terms are on the left hand side.
-2bz-z=a-8
Subtract 8 from both sides.
-2bz=a-8+z
Add z to both sides.
\left(-2z\right)b=z+a-8
The equation is in standard form.
\frac{\left(-2z\right)b}{-2z}=\frac{z+a-8}{-2z}
Divide both sides by -2z.
b=\frac{z+a-8}{-2z}
Dividing by -2z undoes the multiplication by -2z.
b=-\frac{z+a-8}{2z}
Divide z-8+a by -2z.
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