Changing the subject of a linear formula (also called transposing a formula)
Changing the subject of a formula involving a square or square root.
Key ideas
We must always work out what operations were done to the letter that we want to isolate, and then use inverse operations to undo those steps in reverse order.
Whatever we do to one side of the formula, we must do to the other.
An inverse operation just means doing a process backwards.
Non-mathematical example:
Operation: Socks on, then shoes on.
Inverse: Shoes off, then socks off.
Notice how the order of operations has to reverse. You can't take your socks off first!
Mathematical example:
Operation: \(\times 3\), then \(+2\)
Inverse: \(-2\), then \(\div 3\) (in that order)
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Example 1 (non-calculator)
Change the subject of the formula \(y=mx+c\) to \(x\).
Look at the right hand side, where \(x\) is. Think what steps were taken to create \(mx+c\).
First, \(x\) was multiplied by \(m\) to make \(mx\). Then \(c\) was added to make \(mx+c\).
To reverse these steps, we must first subtract \(c\) and then divide by \(m\).
$$
\begin{eqnarray}
y = mx+c\\[6pt]
y-c = mx\\[6pt]
\small\frac{y-c}{m}\normalsize = x\\[6pt]
x = \small\frac{y-c}{m}\normalsize
\end{eqnarray}
$$
Example 2 (non-calculator)
Change the subject of the formula \(e = mc^2\) to \(m\).
Look at the right hand side, where \(m\) is. Think what steps were taken to create \(mc^2\).
There is actually only one step in this example: \(m\) was multiplied by \(c^2\).
So all we have to do here is to divide both sides by \(c^2\).
$$
\begin{eqnarray}
e = mc^2\\[6pt]
\small\frac{e}{c^2}\normalsize = m\\[6pt]
m = \small\frac{e}{c^2}\normalsize
\end{eqnarray}
$$
Example 3 (non-calculator)
Change the subject of the formula \(e = mc^2\) to \(c\).
This looks very similar to the previous example, but don't be fooled!
Look at the right hand side, where \(c\) is. Think what steps were taken to create \(mc^2\).
There are two steps: first square the \(c\) to get \(c^2\) and then multiply by \(m\) to get \(mc^2\).
Again, we have to undo these steps in reverse order: first divide by \(m\) and then square root.
$$
\begin{eqnarray}
e = mc^2\\[6pt]
\small\frac{e}{m}\normalsize = c^2\\[6pt]
\pm\small\sqrt{\frac{e}{m}}\normalsize = c\\[6pt]
c = \pm\small\sqrt{\frac{e}{m}}\normalsize
\end{eqnarray}
$$
The ± sign (plus or minus) is because a negative number squared gives a positive answer, so when you square root both sides, you need to include both possibilities.
Sometimes the context allows us to ignore the ± sign. For example, if we were working with the formula for the area of a circle, \(A=\pi r^2\), then there is no such thing as a negative radius!
Example 4 (non-calculator)
Change the subject of the formula \(v^2=u^2+2as\) to \(a\).
First we subtract \(u^2\) and then divide by \(2s\).
Change the subject of the formula \(a=b \sqrt{c}-d\) to \(c\).
The \(c\) is under a square root sign, so first we make \( \sqrt{c} \) the subject, and then we square both sides.
$$
\begin{eqnarray}
a = b \sqrt{c}-d \\[6pt]
a+d = b \sqrt{c} \\[6pt]
\small\frac{a+d}{b}\normalsize = \sqrt{c} \\[6pt]
\small\left(\frac{a+d}{b}\right)^2\normalsize = c \\[6pt]
c = \small\left(\frac{a+d}{b}\right)^2\normalsize
\end{eqnarray}
$$
It is better not to expand the bracket. The answer is neater in this factorised form, or as:
$$
\begin{eqnarray}
c = \small\frac{(a+d)^2}{b^2}\normalsize
\end{eqnarray}
$$
Example 6 (non-calculator)
Change the subject of the formula \( \large r=\frac{s\,-\,t}{4} \normalsize \) to \(t\small.\)
Our first step here it to multiply through by 4 to remove the fraction.
$$
\begin{eqnarray}
r = \small \frac{s-t}{4}\normalsize \\[6pt]
4r = s-t \\[6pt]
4r-s = -t \\[6pt]
-4r+s = t \\[6pt]
t = -4r+s \\[6pt]
t=s-4r
\end{eqnarray}
$$
As you can see, this example ended up with quite fiddly working, especially at the point when we had \(-t\) on the right and needed to multiply through by -1 to change all the signs.
To avoid this, alternative working is to add \(t\) to both sides after the first step:
$$
\begin{eqnarray}
r = \small \frac{s-t}{4}\normalsize \\[6pt]
4r = s-t \\[6pt]
t+4r = s \\[6pt]
t = s-4r
\end{eqnarray}
$$
Example 7 (non-calculator)
Change the subject of the formula \( ax+bx=c \) to \(x\small.\)
Because the variable \(x\) appears in two terms on the left hand side, our first step is to factorise \(x\) out.
$$
\begin{eqnarray}
ax+bx = c \\[6pt]
x(a+b) = c \\[6pt]
x = \small\frac{c}{a+b}
\end{eqnarray}
$$