Fractional Equations

Problem 212: Solve for $y$ :

(1) $\begin{equation*} \frac{1}{2}y - 2y^{-1} = 0.\end{equation*}$

Solution: Let’s start with writing the term with constant coefficient.

(2) $\begin{equation*} \frac{1}{2}y - 2y^{-1} = \frac{1}{2}y - \frac{2}{y} = 0. \end{equation*}$

One can now find a common denominator in order to add the two fractions. Also, we can multiply by $y$ both sides in order to simplify.

(3) $\begin{align*}y \cdot \left( \frac{1}{2}y - \frac{2}{y} \right) & = ( 0 ) \cdot y \\ \frac{1}{2} y^2 - 2 & = 0 \\ \frac{1}{2}y^2 & = 2 \\ y^2 & = 4 \\ y &= \pm 2.\end{align*}$

That is, $y = \pm 2$ . The last step is to check that this are indeed the solutions to the fractional equations. I will leave that step to you!

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