Solve for x
x=\frac{2+z-3y}{2}
Solve for y
y=\frac{2+z-2x}{3}
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2x-z=2-3y
Subtract 3y from both sides.
2x=2-3y+z
Add z to both sides.
2x=2+z-3y
The equation is in standard form.
\frac{2x}{2}=\frac{2+z-3y}{2}
Divide both sides by 2.
x=\frac{2+z-3y}{2}
Dividing by 2 undoes the multiplication by 2.
x=\frac{z}{2}-\frac{3y}{2}+1
Divide 2-3y+z by 2.
3y-z=2-2x
Subtract 2x from both sides.
3y=2-2x+z
Add z to both sides.
3y=2+z-2x
The equation is in standard form.
\frac{3y}{3}=\frac{2+z-2x}{3}
Divide both sides by 3.
y=\frac{2+z-2x}{3}
Dividing by 3 undoes the multiplication by 3.
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Limits
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