National 5 Maths
Changing the Subject

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 \(m\).

$$ \begin{eqnarray} y = mx+c\\[6pt] y-c = mx\\[6pt] \small\frac{y-c}{m}\normalsize = x\\[6pt] x = \small\frac{y-c}{m}\normalsize \end{eqnarray} $$

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 \(c^2\).

$$ \begin{eqnarray} e = mc^2\\[6pt] \small\frac{e}{c^2}\normalsize = m\\[6pt] m = \small\frac{e}{c^2}\normalsize \end{eqnarray} $$

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!

First we subtract \(u^2\) and then divide by \(2s\).

$$ \begin{eqnarray} v^2 = u^2+2as\\[6pt] v^2-u^2 = 2as\\[6pt] \small\frac{v^2-u^2}{2s}\normalsize = a\\[6pt] a = \small\frac{v^2-u^2}{2s}\normalsize \end{eqnarray} $$

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:

$$ \begin{eqnarray} c = \small\frac{(a+d)^2}{b^2}\normalsize \end{eqnarray} $$

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:

$$ \begin{eqnarray} r = \small \frac{s-t}{4}\normalsize \\[6pt] 4r = s-t \\[6pt] t+4r = s \\[6pt] t = s-4r \end{eqnarray} $$

Because the variable \(x\) appears in two terms on the left hand side, our first step is to factorise \(x\) out.

$$ \begin{eqnarray} ax+bx = c \\[6pt] x(a+b) = c \\[6pt] x = \small\frac{c}{a+b} \end{eqnarray} $$

$$ \begin{eqnarray} s = ut+\small\frac12\normalsize at^2\\[6pt] s-ut = \small\frac12\normalsize at^2\\[6pt] 2(s-ut) = at^2\\[6pt] \small\frac{2(s-ut)}{t^2}\normalsize = a \\[6pt] a = \small\frac{2(s-ut)}{t^2}\normalsize \end{eqnarray} $$

$$ \begin{eqnarray} L = \sqrt{4kt-p}\\[6pt] L^2 = 4kt-p\\[6pt] L^2+p = 4kt\\[6pt] \small\frac{L^2+p}{4t}\normalsize = k\\[6pt] k = \small\frac{L^2+p}{4t}\normalsize \end{eqnarray} $$

$$ \begin{eqnarray} F = \small\frac{t^2+4b}{c}\normalsize\\[6pt] Fc = t^2+4b\\[6pt] Fc-t^2 = 4b\\[6pt] \small\frac{Fc-t^2}{4}\normalsize = b\\[6pt] b = \small\frac{Fc-t^2}{4}\normalsize \end{eqnarray} $$

$$ \begin{eqnarray} y = g\sqrt{x}+h\\[6pt] y-h = g\sqrt{x}\\[6pt] \small\frac{y-h}{g}\normalsize = \sqrt{x}\\[6pt] \small\left(\frac{y-h}{g}\right)^2\normalsize = x\\[6pt] x = \small\left(\frac{y-h}{g}\right)^2\normalsize \end{eqnarray} $$

Method 1:

$$ \begin{eqnarray} A = \small\frac12\normalsize h(x+y)\\[6pt] 2A = h(x+y)\\[6pt] 2A = hx+hy\\[6pt] 2A-hy = hx\\[6pt] \small\frac{2A-hy}{h}\normalsize = x\\[6pt] x = \small\frac{2A-hy}{h}\normalsize \end{eqnarray} $$

Method 2:

$$ \begin{eqnarray} A = \small\frac12\normalsize h(x+y)\\[6pt] 2A = h(x+y)\\[6pt] \small\frac{2A}{h}\normalsize = x+y\\[6pt] \small\frac{2A}{h}\normalsize-y = x\\[6pt] x=\small\frac{2A}{h}\normalsize-y \end{eqnarray} $$

Note: Although the two methods give answers that look different, it can be shown that they are equal.

$$ \begin{eqnarray} d = \sqrt{\frac{3h}{2}}\\[6pt] d^2 = \frac{3h}{2}\\[6pt] 2d^2 = 3h\\[6pt] \small\frac{2d^2}{3}\normalsize = h\\[6pt] h = \small\frac{2d^2}{3}\normalsize \end{eqnarray} $$

$$ \begin{eqnarray} D = \small\frac{B+4}{C^2}\normalsize\\[6pt] C^{2}D = B+4\\[6pt] C^{2}D-4 = B\\[6pt] B=C^{2}D-4 \end{eqnarray} $$

$$ \begin{eqnarray} P = \small\frac13\normalsize mn-r\\[6pt] P+r = \small\frac13\normalsize mn\\[6pt] 3(P+r) = mn\\[6pt] \small\frac{3(P+r)}{n}\normalsize = m \\[6pt] m = \small\frac{3(P+r)}{n}\normalsize \end{eqnarray} $$

Our first step here it to multiply through by \(e\) to remove the fraction.

$$ \begin{eqnarray} f = \small \frac{2d+3}{e}\normalsize \\[6pt] ef = 2d+3 \\[6pt] ef-3 = 2d \\[6pt] 2d = ef-3 \\[6pt] d = \small\frac{ef-3}{2} \\[6pt] \end{eqnarray} $$