Solve for f
\left\{\begin{matrix}f=\frac{-\sin(x+1)+2y}{x}\text{, }&x\neq 0\\f\in \mathrm{R}\text{, }&y=\frac{\sin(1)}{2}\text{ and }x=0\end{matrix}\right.
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-fx=\sin(x+1)-2y
Subtract 2y from both sides.
\left(-x\right)f=\sin(x+1)-2y
The equation is in standard form.
\frac{\left(-x\right)f}{-x}=\frac{\sin(x+1)-2y}{-x}
Divide both sides by -x.
f=\frac{\sin(x+1)-2y}{-x}
Dividing by -x undoes the multiplication by -x.
f=-\frac{\sin(x+1)-2y}{x}
Divide \sin(x+1)-2y by -x.
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