Solve for f
f=\frac{1}{2}+\frac{3}{x}
x\neq 0
Solve for x
x=\frac{6}{2f-1}
f\neq \frac{1}{2}
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fx=\frac{1}{2}x+3
Add 3 to both sides.
xf=\frac{x}{2}+3
The equation is in standard form.
\frac{xf}{x}=\frac{\frac{x}{2}+3}{x}
Divide both sides by x.
f=\frac{\frac{x}{2}+3}{x}
Dividing by x undoes the multiplication by x.
f=\frac{1}{2}+\frac{3}{x}
Divide \frac{x}{2}+3 by x.
fx-3-\frac{1}{2}x=0
Subtract \frac{1}{2}x from both sides.
fx-\frac{1}{2}x=3
Add 3 to both sides. Anything plus zero gives itself.
\left(f-\frac{1}{2}\right)x=3
Combine all terms containing x.
\frac{\left(f-\frac{1}{2}\right)x}{f-\frac{1}{2}}=\frac{3}{f-\frac{1}{2}}
Divide both sides by f-\frac{1}{2}.
x=\frac{3}{f-\frac{1}{2}}
Dividing by f-\frac{1}{2} undoes the multiplication by f-\frac{1}{2}.
x=\frac{6}{2f-1}
Divide 3 by f-\frac{1}{2}.
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