Section 2.4 : Equations With More Than One Variable
4. Solve \(\displaystyle A - \frac{{1 - 2t}}{{4p}} = \frac{{4 + 3t}}{{5p}}\) for \(t\).
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Start SolutionNote that there quite a few solution “paths” that you can take to get the solution to this problem. For this solution let’s first clear the denominator out by multiplying both sides by 20\(p\).
\[\begin{align*}A - \frac{{1 - 2t}}{{4p}} & = \frac{{4 + 3t}}{{5p}}\\ 20p\left( {A - \frac{{1 - 2t}}{{4p}}} \right) & = 20p\left( {\frac{{4 + 3t}}{{5p}}} \right)\\ 20Ap - 5\left( {1 - 2t} \right) & = 4\left( {4 + 3t} \right)\\ 20Ap - 5 + 10t & = 16 + 12t\end{align*}\]We also distributed the constants through the parenthesis in anticipation of the next step.
Show Step 2Now let’s get all the terms with \(t\) on one side and the terms without \(t\) on the other side. Doing this gives,
\[20Ap - 21 = 2t\] Show Step 3Finally, all we need to do is divide by both sides by the coefficient of the \(t\) to get,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{t = \frac{{20Ap - 21}}{2}}}\]Note that depending upon the path you chose for your solution you may have something slightly different for your answer. However, you could do some manipulation of your answer to make it look like mine (or you could manipulate mine to make it look like yours).