Solve for x
x=\frac{y\left(83z+1\right)}{18}
Solve for y
\left\{\begin{matrix}y=\frac{18x}{83z+1}\text{, }&z\neq -\frac{1}{83}\\y\in \mathrm{R}\text{, }&x=0\text{ and }z=-\frac{1}{83}\end{matrix}\right.
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-83yz+18x=y
Swap sides so that all variable terms are on the left hand side.
18x=y+83yz
Add 83yz to both sides.
18x=83yz+y
The equation is in standard form.
\frac{18x}{18}=\frac{83yz+y}{18}
Divide both sides by 18.
x=\frac{83yz+y}{18}
Dividing by 18 undoes the multiplication by 18.
x=\frac{y\left(83z+1\right)}{18}
Divide y+83yz by 18.
y+83yz=18x
Add 83yz to both sides.
\left(1+83z\right)y=18x
Combine all terms containing y.
\left(83z+1\right)y=18x
The equation is in standard form.
\frac{\left(83z+1\right)y}{83z+1}=\frac{18x}{83z+1}
Divide both sides by 83z+1.
y=\frac{18x}{83z+1}
Dividing by 83z+1 undoes the multiplication by 83z+1.
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