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Section 2.2 : Linear Equations

4. Solve the following equation and check your answer.

\[\frac{{4t}}{{{t^2} - 25}} = \frac{1}{{5 - t}}\]

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Hint : Do not forget to watch out for values of \(t\) that we’ll need to avoid!
Start Solution

Let’s first factor the denominator on the left side so we can identify the LCD. While we are at it we will also factor a minus out of the denominator on the right side.

\[\begin{align*}\frac{{4t}}{{{t^2} - 25}} & = \frac{1}{{5 - t}}\\ \frac{{4t}}{{\left( {t - 5} \right)\left( {t + 5} \right)}} & = \frac{1}{{ - \left( {t - 5} \right)}}\\ \frac{{4t}}{{\left( {t - 5} \right)\left( {t + 5} \right)}} & = - \frac{1}{{t - 5}}\end{align*}\]

So, after factoring the left side and factoring the minus sign out of the denominator on the right side we can quickly see that the LCD for this equation is,

\[\left( {t - 5} \right)\left( {t + 5} \right)\]

From this we can also see that we’ll need to avoid \(t = 5\) and \(t = - 5\). Remember that we have to avoid division by zero and we will clearly get division by zero with each of these values of \(t\).

Show Step 2

Next, we need to do find the solution. To get the solution we’ll need to multiply both sides by the LCD and the go through the same process we used in the first couple of practice problems. Here is that work.

\[\begin{align*}\left( {t - 5} \right)\left( {t + 5} \right)\left( {\frac{{4t}}{{\left( {t - 5} \right)\left( {t + 5} \right)}}} \right) & = - \left( {\frac{1}{{t - 5}}} \right)\left( {t - 5} \right)\left( {t + 5} \right)\\ 4t & = - \left( {t + 5} \right)\\ 4t & = - t - 5\\ 5t & = - 5\\ t & = - 1\end{align*}\] Show Step 3

Finally, we need to verify that our answer from Step 2 is in fact a solution.

The first thing to note is that it is not one of the values of \(t\) that we need to avoid. Having determined that we know that we do have a potential solution (i.e. it’s not a value of \(t\) we need to avoid) all we need to do is plug the solution into the equation given in the problem statement.

Here is the verification work.

\[\begin{align*}\frac{{4\left( { - 1} \right)}}{{{{\left( { - 1} \right)}^2} - 25}} & \mathop = \limits^? \frac{1}{{5 - \left( { - 1} \right)}}\\ \frac{{ - 4}}{{1 - 25}} & \mathop = \limits^? \frac{1}{{5 + 1}}\\ \frac{1}{6} & = \frac{1}{6}\hspace{0.5in} {\mbox{OK}}\end{align*}\]

So, we can see that our solution from Step 2 is in fact the solution to the equation.