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