Solve for x
x=-\left(y-7z\right)
Solve for y
y=-\left(x-7z\right)
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\frac{1}{7}x+\frac{1}{7}y=z
Divide each term of x+y by 7 to get \frac{1}{7}x+\frac{1}{7}y.
\frac{1}{7}x=z-\frac{1}{7}y
Subtract \frac{1}{7}y from both sides.
\frac{1}{7}x=-\frac{y}{7}+z
The equation is in standard form.
\frac{\frac{1}{7}x}{\frac{1}{7}}=\frac{-\frac{y}{7}+z}{\frac{1}{7}}
Multiply both sides by 7.
x=\frac{-\frac{y}{7}+z}{\frac{1}{7}}
Dividing by \frac{1}{7} undoes the multiplication by \frac{1}{7}.
x=7z-y
Divide z-\frac{y}{7} by \frac{1}{7} by multiplying z-\frac{y}{7} by the reciprocal of \frac{1}{7}.
\frac{1}{7}x+\frac{1}{7}y=z
Divide each term of x+y by 7 to get \frac{1}{7}x+\frac{1}{7}y.
\frac{1}{7}y=z-\frac{1}{7}x
Subtract \frac{1}{7}x from both sides.
\frac{1}{7}y=-\frac{x}{7}+z
The equation is in standard form.
\frac{\frac{1}{7}y}{\frac{1}{7}}=\frac{-\frac{x}{7}+z}{\frac{1}{7}}
Multiply both sides by 7.
y=\frac{-\frac{x}{7}+z}{\frac{1}{7}}
Dividing by \frac{1}{7} undoes the multiplication by \frac{1}{7}.
y=7z-x
Divide z-\frac{x}{7} by \frac{1}{7} by multiplying z-\frac{x}{7} by the reciprocal of \frac{1}{7}.
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