Solve for x
x=3+\frac{7}{2y}
y\neq 0
Solve for y
y=-\frac{7}{2\left(3-x\right)}
x\neq 3
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2xy-7=6y
Multiply both sides of the equation by y.
2xy=6y+7
Add 7 to both sides.
2yx=6y+7
The equation is in standard form.
\frac{2yx}{2y}=\frac{6y+7}{2y}
Divide both sides by 2y.
x=\frac{6y+7}{2y}
Dividing by 2y undoes the multiplication by 2y.
x=3+\frac{7}{2y}
Divide 6y+7 by 2y.
2xy-7=6y
Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.
2xy-7-6y=0
Subtract 6y from both sides.
2xy-6y=7
Add 7 to both sides. Anything plus zero gives itself.
\left(2x-6\right)y=7
Combine all terms containing y.
\frac{\left(2x-6\right)y}{2x-6}=\frac{7}{2x-6}
Divide both sides by 2x-6.
y=\frac{7}{2x-6}
Dividing by 2x-6 undoes the multiplication by 2x-6.
y=\frac{7}{2\left(x-3\right)}
Divide 7 by 2x-6.
y=\frac{7}{2\left(x-3\right)}\text{, }y\neq 0
Variable y cannot be equal to 0.
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