Fractional Equations II

Problem 213: Solve for $t$ :

(1) $\begin{equation*} \frac{t}{t-6} + \frac{12}{t^2 - 6t} = - \frac{4}{t}\end{equation*}$

Solution: We will start by factoring the denominators.

(2) $\begin{align*} \frac{t}{t-6} + \frac{12}{t^2 - 6t} = \frac{t}{t-6} + \frac{12}{t(t - 6)} = - \frac{4}{t} \end{align*}$

Now we get the denominator for the fractions so that we can add them. That is,

(3) $\begin{align*}\frac{t}{t-6} + \frac{12}{t(t - 6)} &= - \frac{4}{t} \\ \frac{t \cdot t}{t \cdot (t-6)} + \frac{12}{t(t - 6)} & = - \frac{4}{t} \\ \frac{t^2}{t(t-6)} + \frac{12}{t(t - 6)} & = - \frac{4}{t} \\ \frac{t^2 + 12}{t(t-6)} & = - \frac{4}{t}. \end{align*}$

Now we can get rid of the denominator by multiplying both side by it.

(4) $\begin{align*}t(t-6) \cdot \left( \frac{t^2 + 12}{t(t-6)} \right) & = t(t-6) \cdot \left( - \frac{4}{t}\right) \\ t^2 + 12 & = -4(t-6) \\ t^2 + 12 & = -4t + 24 \\ t^2 + 4t - 12 & = 0 \\ (t + 6)(t-2) & = 0 \\ & \Rightarrow t = -6, t = 2. \end{align*}$

That is, $t = -6$ and $t = 2$ . The last step is to check that these are indeed the solutions by plugging in back the values. I will leave that to you!

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