Solve for x
x=\frac{1-3y}{4}
Solve for y
y=\frac{1-4x}{3}
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\frac{1}{3}x=\frac{1}{12}-\frac{1}{4}y
Subtract \frac{1}{4}y from both sides.
\frac{1}{3}x=-\frac{y}{4}+\frac{1}{12}
The equation is in standard form.
\frac{\frac{1}{3}x}{\frac{1}{3}}=\frac{-\frac{y}{4}+\frac{1}{12}}{\frac{1}{3}}
Multiply both sides by 3.
x=\frac{-\frac{y}{4}+\frac{1}{12}}{\frac{1}{3}}
Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.
x=\frac{1-3y}{4}
Divide \frac{1}{12}-\frac{y}{4} by \frac{1}{3} by multiplying \frac{1}{12}-\frac{y}{4} by the reciprocal of \frac{1}{3}.
\frac{1}{4}y=\frac{1}{12}-\frac{1}{3}x
Subtract \frac{1}{3}x from both sides.
\frac{1}{4}y=-\frac{x}{3}+\frac{1}{12}
The equation is in standard form.
\frac{\frac{1}{4}y}{\frac{1}{4}}=\frac{-\frac{x}{3}+\frac{1}{12}}{\frac{1}{4}}
Multiply both sides by 4.
y=\frac{-\frac{x}{3}+\frac{1}{12}}{\frac{1}{4}}
Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.
y=\frac{1-4x}{3}
Divide \frac{1}{12}-\frac{x}{3} by \frac{1}{4} by multiplying \frac{1}{12}-\frac{x}{3} by the reciprocal of \frac{1}{4}.
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