Solve for a
a=\frac{n}{2}+\frac{1}{2b}
b\neq 0
Solve for b
b=-\frac{1}{n-2a}
n\neq 2a
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2ba=bn+1
Swap sides so that all variable terms are on the left hand side.
\frac{2ba}{2b}=\frac{bn+1}{2b}
Divide both sides by 2b.
a=\frac{bn+1}{2b}
Dividing by 2b undoes the multiplication by 2b.
a=\frac{n}{2}+\frac{1}{2b}
Divide bn+1 by 2b.
bn+1-2ba=0
Subtract 2ba from both sides.
bn-2ba=-1
Subtract 1 from both sides. Anything subtracted from zero gives its negation.
\left(n-2a\right)b=-1
Combine all terms containing b.
\frac{\left(n-2a\right)b}{n-2a}=-\frac{1}{n-2a}
Divide both sides by n-2a.
b=-\frac{1}{n-2a}
Dividing by n-2a undoes the multiplication by n-2a.
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