Transposition of formulas – Brackets and fractions
Expand your transposition skills by exploring how to rearrange formulas involving brackets and fractions. By doing this, you will develop the skills needed to isolate variables in more complex expressions and prepare yourself well for advanced problem-solving in STEM.
You will have learned the necessary theory when you were first introduced to transposition of formulas, so this page focuses on examples.
Example 1 – transposing formulas with brackets and fractions
The formula to calculate simple interest is:
\[A=P(1+rn)\]
where \(A\) is the total amount, \(P\) is the principal or initial amount, \(r\) is the interest rate and \(n\) is the time or duration.
Express the formula in a way that helps calculate the time taken to generate an amount of interest.
We need to undo the operations to get \(n\) on its own on one side of the equal sign. On the right, \(n\) is multiplied by \(r\), added to \(1\), then multiplied by \(P\). To undo this, we need to work backwards and divide by \(P\) first, then subtract \(1\), and finally, divide by \(r\).
\[\begin{align*} A & = P(1+rn)\quad\textrm{divide by \(P\) on both sides}\\
\frac{A}{P} & = \frac{P(1+rn)}{P}\quad\textrm{cancel out \(P\) on right}\\
\frac{A}{P} & = 1+rn\quad\textrm{subtract \(1\) on both sides}\\
\frac{A}{P}-1 & = 1+rn-1\\
\frac{A}{P}-1 & = rn\quad\textrm{divide by \(r\) on both sides}\\
\frac{\frac{A}{P}-1}{r} & = \frac{rn}{r}\quad\textrm{cancel out \(r\) on right}\\
\frac{\frac{A}{P}-1}{r} & = n
\end{align*}\]
The fraction on the left is a bit messy, so we can rewrite it in a different way. Remember that dividing by something is the same as multiplying by its reciprocal. We can rewrite \(\div r\) as \(\times \dfrac{1}{r}\). So:
This can be rewritten with \(n\) on the left. We can also remove the \(\times\) sign, since when we use brackets, it is assumed that any term outside the bracket is multiplied by whatever is inside the brackets.
\[n=\frac{1}{r}\left(\frac{A}{P}-1\right)\]
Transform the formula \(P=2(L-W)\) to make \(L\) the subject.
We need to undo the operations to get \(L\) on its own on one side of the equal sign. On the right, \(W\) is subtracted from \(L\), then this is multiplied by \(2\). To undo this, we need to work backwards and divide by \(2\) first, then add \(W\).
\[\begin{align*} P & = 2(L-W)\quad\textrm{divide both sides by \(2\)}\\
\frac{P}{2} & = \frac{2(L-W)}{2}\quad\textrm{cancel out \(2\) on right}\\
\frac{P}{2} & = L-W\quad\textrm{add \(W\) to both sides}\\
\frac{P}{2}+W & = L-W+W\\
\frac{P}{2}+W & = L
\end{align*}\]
We can rewrite this with \(L\) on the left.
\[L=\frac{P}{2}+W\]
Transform the formula \(P=2(L-W)\) to make \(W\) the subject.
We need to undo the operations to get \(W\) on its own on one side of the equal sign. On the right, \(W\) is subtracted from \(L\), then this is multiplied by \(2\). To undo this, we need to work backwards and divide by \(2\) first, then subtract \(L\) and undo the negative in front of the \(W\).
\[\begin{align*} P & = 2(L-W)\quad\textrm{divide both sides by \(2\)}\\
\frac{P}{2} & = \frac{2(L-W)}{2}\quad\textrm{cancel out \(2\) on right}\\
\frac{P}{2} & = L-W\quad\textrm{subtract \(L\) from both sides}\\
\frac{P}{2}-L & = L-W-L\\
\frac{P}{2}-L & = -W\quad\textrm{invert all signs to undo negative}\\
-\frac{P}{2}+L & = W\\
L-\frac{P}{2} & = W
\end{align*}\]
We can rewrite this with \(W\) on the left.
\[W = L-\frac{P}{2}\]
Rearrange the formula \(L=\dfrac{Mt-g}{b}\) to make \(M\) the subject.
We need to undo the operations to get \(M\) on its own on one side of the equal sign. On the right, \(M\) is multiplied by \(t\), then \(g\) is subtracted and the whole expression is divided by \(b\). To undo this, we need to work backwards and multiply by \(b\) first, add \(g\), then divide by \(t\).
\[\begin{align*} L & = \frac{Mt-g}{b}\quad\textrm{multiply both sides by \(b\)}\\
L\times b & = \frac{Mt-g}{b} \times b\quad\textrm{cancel out \(b\) on right}\\
Lb & = Mt-g\quad\textrm{add \(g\) on both sides}\\
Lb +g & = Mt-g+g\\
Lb +g & = Mt\quad\textrm{divide by \(t\) on both sides}\\
\frac{Lb +g}{t} & = M
\end{align*}\]
We can rewrite this with \(M\) on the left.
\[M=\frac{Lb +g}{t}\]
Make \(v\) the subject of \(E=mgh+\dfrac{1}{2}mv^{2}\).
We need to undo the operations to get \(v\) on its own on one side of the equal sign. On the right, \(v\) is squared, multiplied by \(\dfrac{1}{2}m\), then added to \(mgh\). To undo this, we need to work backwards and subtract \(mgh\), divide by \(\dfrac{1}{2}m\), then take the square root. For the division step, you could also multiply by \(2\) since this is the same as dividing by \(\dfrac{1}{2}\).
For this particular worked example, not all of the working steps are shown. This is the level of working you should aim for.
\[\begin{align*} E & = mgh+\frac{1}{2}mv^{2}\quad\textrm{subtract \(mgh\) from both sides}\\
E-mgh & = \frac{1}{2}mv^{2}\quad\textrm{multiply both sides by \(2\)}\\
2(E-mgh) & = mv^{2}\quad\textrm{divide by \(m\)}\\
\frac{2(E-mgh)}{m} & = v^{2}\quad\textrm{take square root of both sides}\\
\pm \sqrt{\frac{2(E-mgh)}{m}} & = v
\end{align*}\]
Since the formulas you work with will most likely relate to a physical situation, a negative solution may not make sense. Therefore, we can just take the positive solution as the final answer.
\[v=\sqrt{\frac{2(E-mgh)}{m}}\]
If \(\dfrac{2}{k}=\dfrac{j+1}{3}\), find \(k\).
We need to undo the operations to get \(k\) on its own on one side of the equal sign. On the left, \(2\) is divided by \(k\). To undo this, we need to multiply by \(k\), then rearrange to isolate \(k\).
\[\begin{align*} \frac{2}{k} & = \frac{j+1}{3}\quad\textrm{multiply both sides by \(k\)}\\
2 & = \frac{k(j+1)}{3}\quad\textrm{multiply both sides by \(3\)}\\
6 & = k(j+1)\quad\textrm{divide both sides by \((j+1)\)}\\
\frac{6}{j+1} & = k
\end{align*}\]
This process is also called cross-multiplication where the opposing numerators and denominators are multiplied.
We can rewrite the final answer with \(k\) on the left.
\[k=\frac{6}{j+1}\]
Your turn – transposing formulas with brackets and fractions
