Solve for x
\left\{\begin{matrix}x=-\frac{4y}{2-3z}\text{, }&z\neq \frac{2}{3}\\x\in \mathrm{R}\text{, }&y=0\text{ and }z=\frac{2}{3}\end{matrix}\right.
Solve for y
y=\frac{x\left(3z-2\right)}{4}
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2x+4y-3zx=0
Subtract 3zx from both sides.
2x-3zx=-4y
Subtract 4y from both sides. Anything subtracted from zero gives its negation.
\left(2-3z\right)x=-4y
Combine all terms containing x.
\frac{\left(2-3z\right)x}{2-3z}=-\frac{4y}{2-3z}
Divide both sides by 2-3z.
x=-\frac{4y}{2-3z}
Dividing by 2-3z undoes the multiplication by 2-3z.
4y=3zx-2x
Subtract 2x from both sides.
4y=3xz-2x
The equation is in standard form.
\frac{4y}{4}=\frac{x\left(3z-2\right)}{4}
Divide both sides by 4.
y=\frac{x\left(3z-2\right)}{4}
Dividing by 4 undoes the multiplication by 4.
y=\frac{3xz}{4}-\frac{x}{2}
Divide x\left(-2+3z\right) by 4.
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