Solve for x
x=\frac{1}{4}+\frac{3}{8y}
y\neq 0
Solve for y
y=\frac{3}{2\left(4x-1\right)}
x\neq \frac{1}{4}
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8xy=3+2y
Add 2y to both sides.
8yx=2y+3
The equation is in standard form.
\frac{8yx}{8y}=\frac{2y+3}{8y}
Divide both sides by 8y.
x=\frac{2y+3}{8y}
Dividing by 8y undoes the multiplication by 8y.
x=\frac{1}{4}+\frac{3}{8y}
Divide 3+2y by 8y.
\left(8x-2\right)y=3
Combine all terms containing y.
\frac{\left(8x-2\right)y}{8x-2}=\frac{3}{8x-2}
Divide both sides by 8x-2.
y=\frac{3}{8x-2}
Dividing by 8x-2 undoes the multiplication by 8x-2.
y=\frac{3}{2\left(4x-1\right)}
Divide 3 by 8x-2.
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