Solve for x
x=\frac{7y-z}{3}
Solve for y
y=\frac{3x+z}{7}
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15x+5z=35y
Add 35y to both sides. Anything plus zero gives itself.
15x=35y-5z
Subtract 5z from both sides.
\frac{15x}{15}=\frac{35y-5z}{15}
Divide both sides by 15.
x=\frac{35y-5z}{15}
Dividing by 15 undoes the multiplication by 15.
x=\frac{7y-z}{3}
Divide 35y-5z by 15.
-35y+5z=-15x
Subtract 15x from both sides. Anything subtracted from zero gives its negation.
-35y=-15x-5z
Subtract 5z from both sides.
\frac{-35y}{-35}=\frac{-15x-5z}{-35}
Divide both sides by -35.
y=\frac{-15x-5z}{-35}
Dividing by -35 undoes the multiplication by -35.
y=\frac{3x+z}{7}
Divide -15x-5z by -35.
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Limits
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