National 5 Maths
Changing the Subject

Look at the right hand side, where \(x\) is. Think what steps were taken to create \(mx+c\small.\)

First, \(x\) was multiplied by \(m\) to make \(mx\small.\) Then \(c\) was added to make \(mx+c\small.\)

To reverse these steps, we must first subtract \(c\) and then divide by \(m\small.\)

$$ \begin{gather} \underset{\color{red}{\mathsf{-c}}}{y} = \underset{\color{red}{\mathsf{-c}}}{\vphantom{y}mx+c} \\[10pt] \underset{\color{red}{\mathsf{\div m}}}{y-c} = \underset{\color{red}{\mathsf{\div m}}}{\vphantom{y}mx} \\[10pt] \frac{y-c}{m} = x \\[6pt] x = \frac{y-c}{m} \end{gather} $$

Note: We simply swapped sides at the end, so that the formula starts with \(x=\small\cdots\,\small.\) Some teachers prefer to swap sides at the start of the working. Feel free to do it that way, if you prefer.

Video solution by Clelland Maths 

Look at the right hand side, where \(m\) is. Think what steps were taken to create \(mc^2\small.\)

There is actually only one step in this example: \(m\) was multiplied by \(c^2\small.\)

So all we have to do here is to divide both sides by \(c^2\small.\)

$$ \begin{gather} \underset{\color{red}{\mathsf{\div c^2}}}{e} = \underset{\color{red}{\mathsf{\div c^2}}}{mc^2} \\[10pt] \frac{e}{c^2} = m \\[6pt] m = \frac{e}{c^2} \end{gather} $$

Video solution by Clelland Maths 

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.

$$ \begin{gather} \underset{\color{red}{\mathsf{\div m}}}{e} = \underset{\color{red}{\mathsf{\div m}}}{\vphantom{e}mc^2} \\[10pt] \underset{\color{red}{\mathsf{\sqrt{\phantom{x}\:}}}}{\frac{e}{m}} = \underset{\color{red}{\mathsf{\sqrt{\phantom{x}\:}}}}{\vphantom{\frac{e}{m}}c^2} \\[12pt] \pm\,\sqrt{\!\frac{e}{m}\,} = c \\[6pt] c = \pm\,\sqrt{\!\frac{e}{m}\,} \end{gather} $$

Note 1: The \(\pm\) sign (plus or minus) is because a negative number squared gives a positive answer, so when you square root both sides, you should include both possibilities.

Note 2: Sometimes the context allows us to ignore the \(\pm\) sign. For example, if we were working with the formula for the area of a circle, \(A=\pi r^2\small,\) then there is no such thing as a negative radius!

Video solution by Clelland Maths 

The inverse operations are: first subtract \(u^2\) and then divide by \(2s\small.\)

$$ \begin{gather} \underset{\color{red}{\mathsf{-u^2}}}{v^2} = \underset{\color{red}{\mathsf{-u^2}}}{\vphantom{v^2}u^2+2as} \\[10pt] \underset{\color{red}{\mathsf{\div 2s}}}{v^2-u^2} = \underset{\color{red}{\mathsf{\div 2s}}}{\vphantom{v^2-u^2}2as} \\[10pt] \frac{v^2-u^2}{2s} = a \\[6pt] a = \frac{v^2-u^2}{2s} \end{gather} $$

Video solution by Clelland Maths 

The \(c\) is under a square root sign, so first we make \( \sqrt{c\,} \) the subject, and then we square both sides.

$$ \begin{gather} \underset{\color{red}{\mathsf{+d}}}{\vphantom{\sqrt{c}}a} = \underset{\color{red}{\mathsf{+d}}}{b\,\sqrt{c\,}-d} \\[10pt] \underset{\color{red}{\mathsf{\div b}}}{\vphantom{\sqrt{c}}a+d} = \underset{\color{red}{\mathsf{\div b}}}{b\,\sqrt{c\,}} \\[10pt] \underset{\color{red}{\mathsf{(\dots)^2}}}{\frac{a+d}{b}} = \underset{\color{red}{\mathsf{(\dots)^2}}}{\vphantom{\frac{a+d}{b}}\sqrt{c\,}} \\[12pt] \left(\frac{a+d}{b}\right)^2 = c \\[6pt] c = \left(\frac{a+d}{b}\right)^2 \end{gather} $$

Note: It is better not to expand the bracket. The answer is neater in this factorised form, or as:

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

Video solution by Clelland Maths 

Our first step here it to multiply through by 4 to remove the fraction.

$$ \begin{gather} \underset{\color{red}{\mathsf{\small{\times}\normalsize 4}}}{\vphantom{\frac{s-t}{4}}r} = \underset{\color{red}{\mathsf{\small{\times}\normalsize 4}}}{\frac{s-t}{4}} \\[14pt] \underset{\color{red}{\mathsf{-s}}}{4r} = \underset{\color{red}{\mathsf{-s}}}{\vphantom{4r}s-t} \\[10pt] \underset{\color{red}{\mathsf{\small{\times}\normalsize -1}}}{4r-s} = \underset{\color{red}{\mathsf{\small{\times}\normalsize -1}}}{\vphantom{4r-s}-t} \\[10pt] -4r+s = t \\[8pt] t = -4r+s \\[8pt] t=s-4r \end{gather} $$

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:

$$ \begin{gather} \underset{\color{red}{\mathsf{\small{\times}\normalsize 4}}}{\vphantom{\frac{s-t}{4}}r} = \underset{\color{red}{\mathsf{\small{\times}\normalsize 4}}}{\frac{s-t}{4}} \\[14pt] \underset{\color{red}{\mathsf{+t}}}{4r} = \underset{\color{red}{\mathsf{+t}}}{\vphantom{4r}s-t} \\[10pt] \underset{\color{red}{\mathsf{-4r}}}{t+4r} = \underset{\color{red}{\mathsf{-4r}}}{\vphantom{t+4r}s} \\[10pt] t = s-4r \end{gather} $$

Video solution by Clelland Maths 

\(x\) appears twice on the left hand side, so our first step is to factorise \(x\) out.

Factorising will make \(x\) appear only once.

Then we can divide both sides by the other factor, \((a+b)\), like this:

$$ \begin{gather} ax+bx = c \\[10pt] \underset{\color{red}{\mathsf{\div(a+b)}}}{x(a+b)} = \underset{\color{red}{\mathsf{\div(a+b)}}}{\vphantom{x(a+b)}c} \\[9pt] x = \frac{c}{a+b} \end{gather} $$

Note: A question requiring factorisation hasn't yet appeared on an N5 exam paper, but it could!

Video solution by Clelland Maths 

\(a\) has been multiplied by \(t^2\small,\) halved and finally \(ut\) was added.

So the inverse operations are as follows:

$$ \begin{gather} \underset{\color{red}{\mathsf{-ut}}}{\vphantom{\frac{1}{2}}s} = \underset{\color{red}{\mathsf{-ut}}}{ut+\frac{1}{2}at^2} \\[14pt] \underset{\color{red}{\mathsf{\small{\times}\normalsize 2}}}{\vphantom{\frac{1}{2}}s-ut} = \underset{\color{red}{\mathsf{\small{\times}\normalsize 2}}}{\frac{1}{2}at^2} \\[14pt] \underset{\color{red}{\mathsf{\div t^2}}}{2(s-ut)} = \underset{\color{red}{\mathsf{\div t^2}}}{\vphantom{2(s-ut)}at^2} \\[10pt] \frac{2(s-ut)}{t^2} = a \\[6pt] a = \frac{2(s-ut)}{t^2} \end{gather} $$

Note: You could also double the whole equation at the start, giving \(2s=2ut+at^{2}\small,\) before subtracting \(2ut\) from both sides and finally dividing by \(t^{2}\small.\) That's just a matter of personal preference.

Video solution by Clelland Maths 

Because the whole of the right hand side is under a square root sign, our first step is to square both sides:

$$ \begin{gather} \underset{\color{red}{\mathsf{(\dots)^2}}}{\vphantom{\sqrt{4kt-p\,}}L} = \underset{\color{red}{\mathsf{(\dots)^2}}}{\sqrt{4kt-p\,}} \\[14pt] \underset{\color{red}{\mathsf{+p}}}{\vphantom{4kt-p}L^2} = \underset{\color{red}{\mathsf{+p}}}{4kt-p} \\[10pt] \underset{\color{red}{\mathsf{\div 4t}}}{\vphantom{4kt}L^2+p} = \underset{\color{red}{\mathsf{\div 4t}}}{4kt} \\[10pt] \frac{L^2+p}{4t} = k \\[6pt] k = \frac{L^2+p}{4t} \end{gather} $$

Video solution by Clelland Maths 

This is a 3-step solution. Our first job is to get rid of the fraction, by multiplying by its denominator:

$$ \begin{gather} \underset{\color{red}{\mathsf{\small{\times}\normalsize c}}}{\vphantom{\frac{t^2+4b}{c}}F} = \underset{\color{red}{\mathsf{\small{\times}\normalsize c}}}{\frac{t^2+4b}{c}} \\[14pt] \underset{\color{red}{\mathsf{-t^2}}}{Fc} = \underset{\color{red}{\mathsf{-t^2}}}{\vphantom{Fc}t^2+4b} \\[10pt] \underset{\color{red}{\mathsf{\div 4}}}{Fc-t^2} = \underset{\color{red}{\mathsf{\div 4}}}{\vphantom{Fc-t^2}4b} \\[10pt] \frac{Fc-t^2}{4} = b \\[6pt] b = \frac{Fc-t^2}{4} \end{gather} $$

Video solution by Clelland Maths 

\(x\) has been square rooted, then multiplied by \(g\) and finally, \(h\) was added.

So the inverse operations are: subtract \(h\small,\) then divide by \(g\) and finally, square both sides.

$$ \begin{gather} \underset{\color{red}{\mathsf{-h}}}{\vphantom{g\sqrt{x}}y} = \underset{\color{red}{\mathsf{-h}}}{g\,\sqrt{x\,}+h} \\[10pt] \underset{\color{red}{\mathsf{\div g}}}{\vphantom{\sqrt{x}}y-h} = \underset{\color{red}{\mathsf{\div g}}}{g\,\sqrt{x\,}} \\[10pt] \underset{\color{red}{\mathsf{(\dots)^2}}}{\frac{y-h}{g}} = \underset{\color{red}{\mathsf{(\dots)^2}}}{\vphantom{\frac{y-h}{g}}\sqrt{x\,}} \\[14pt] \left(\frac{y-h}{g}\right)^2 = x \\[6pt] x = \left(\frac{y-h}{g}\right)^2 \end{gather} $$

Note: It is better not to expand the bracket. The answer is neater in this factorised form, or as:

$$ \begin{eqnarray} x = \frac{(y-h)^2}{g^2} \end{eqnarray} $$

Video solution by Clelland Maths 

There are two reasonable ways of approaching this question. Both begin by doubling both sides.

Method 1: Expand the bracket

$$ \begin{gather} \underset{\color{red}{\mathsf{\small{\times}\normalsize 2}}}{\vphantom{\frac{1}{2}h(x+y)}A} = \underset{\color{red}{\mathsf{\small{\times}\normalsize 2}}}{\frac{1}{2}h(x+y)} \\[14pt] \underset{\color{red}{\mathsf{expand}}}{\vphantom{h(x+y)}2A} = \underset{\color{red}{\mathsf{expand}}}{h(x+y)} \\[10pt] \underset{\color{red}{\mathsf{-hy}}}{\vphantom{hx+hy}2A} = \underset{\color{red}{\mathsf{-hy}}}{hx+hy} \\[10pt] \underset{\color{red}{\mathsf{\div h}}}{2A-hy} = \underset{\color{red}{\mathsf{\div h}}}{\vphantom{2A-hy}hx} \\[10pt] \frac{2A-hy}{h} = x \\[6pt] x = \frac{2A-hy}{h} \end{gather} $$

Method 2: Just divide by h

$$ \begin{gather} \underset{\color{red}{\mathsf{\small{\times}\normalsize 2}}}{\vphantom{\frac{1}{2}h(x+y)}A} = \underset{\color{red}{\mathsf{\small{\times}\normalsize 2}}}{\frac{1}{2}h(x+y)} \\[14pt] \underset{\color{red}{\mathsf{\div h}}}{\vphantom{h(x+y)}2A} = \underset{\color{red}{\mathsf{\div h}}}{h(x+y)} \\[10pt] \underset{\color{red}{\mathsf{-y}}}{\frac{2A}{h}} = \underset{\color{red}{\mathsf{-y}}}{\vphantom{\frac{2A}{h}}x+y} \\[14pt] \frac{2A}{h}-y = x \\[6pt] x = \frac{2A}{h}-y \end{gather} $$

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

Video solution by Clelland Maths 

Here, \(h\) has been multiplied by \(3\) and divided by \(2\small,\) then square rooted.

So we apply the relevant inverse operations :

$$ \begin{gather} \underset{\color{red}{\mathsf{(\dots)^2}}}{\vphantom{\sqrt{\!\frac{3h}{2}\,}}d} = \underset{\color{red}{\mathsf{(\dots)^2}}}{\sqrt{\frac{3h}{2}}} \\[16pt] \underset{\color{red}{\mathsf{\small{\times}\normalsize 2}}}{\vphantom{\frac{3h}{2}}d^2} = \underset{\color{red}{\mathsf{\small{\times}\normalsize 2}}}{\frac{3h}{2}} \\[14pt] \underset{\color{red}{\mathsf{\div 3}}}{2d^2} = \underset{\color{red}{\mathsf{\div 3}}}{\vphantom{2d^2}3h} \\[10pt] \frac{2d^2}{3} = h \\[6pt] h = \frac{2d^2}{3} \end{gather} $$

Note: Swapping the order of the \(\times 2\) and \(\div 3\) steps also works, as division and multiplication have the same BODMAS/BIDMAS priority.

Video solution by Clelland Maths 

This fairly simple 2-mark rearrangement starts with multiplying both sides by the denominator:

$$ \begin{gather} \underset{\color{red}{\mathsf{\small{\times}\normalsize C^2}}}{\vphantom{\frac{B+4}{C^2}}D} = \underset{\color{red}{\mathsf{\small{\times}\normalsize C^2}}}{\frac{B+4}{C^2}} \\[16pt] \underset{\color{red}{\mathsf{-4}}}{C^2D} = \underset{\color{red}{\mathsf{-4}}}{\vphantom{C^2D}B+4} \\[10pt] C^2D-4 = B \\[8pt] B = C^2D-4 \end{gather} $$

Video solution by Clelland Maths 

This 3-mark, 3-step rearrangement can be done in more than one way.

As an alternative to the working below, you could start by multiplying both sides by \(3\small,\) if you prefer.

$$ \begin{gather} \underset{\color{red}{\mathsf{+r}}}{\vphantom{\frac{1}{3}}P} = \underset{\color{red}{\mathsf{+r}}}{\frac{1}{3}mn-r} \\[16pt] \underset{\color{red}{\mathsf{\small{\times}\normalsize 3}}}{\vphantom{\frac{1}{3}}P+r} = \underset{\color{red}{\mathsf{\small{\times}\normalsize 3}}}{\frac{1}{3}mn} \\[16pt] \underset{\color{red}{\mathsf{\div n}}}{3(P+r)} = \underset{\color{red}{\mathsf{\div n}}}{\vphantom{3(P+r)}mn} \\[10pt] \frac{3(P+r)}{n} = m \\[8pt] m = \frac{3(P+r)}{n} \end{gather} $$

Note: It doesn't matter whether you write the numerator as \(3(P+r)\) or \(3P+3r\small.\)

Video solution by Clelland Maths 

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

$$ \begin{gather} \underset{\color{red}{\mathsf{\small{\times}\normalsize e}}}{\vphantom{\frac{2d+3}{e}}f} = \underset{\color{red}{\mathsf{\small{\times}\normalsize e}}}{\frac{2d+3}{e}} \\[16pt] \underset{\color{red}{\mathsf{-3}}}{ef} = \underset{\color{red}{\mathsf{-3}}}{\vphantom{ef}2d+3} \\[10pt] \underset{\color{red}{\mathsf{swap}}}{ef-3} = \underset{\color{red}{\mathsf{swap}}}{\vphantom{ef-3}2d} \\[10pt] \underset{\color{red}{\mathsf{\div 2}}}{\vphantom{ef}2d} = \underset{\color{red}{\mathsf{\div 2}}}{ef-3} \\[10pt] d = \frac{ef-3}{2} \end{gather} $$

Note: You will have noticed that we swapped sides in the middle of the working above. Usually we swap at the end. Some people prefer to swap at the very start. It really doesn't matter!

Video solution by Clelland Maths 

This is quite a fiddly 3-mark, 3-step rearrangement:

$$ \begin{gather} \underset{\color{red}{\mathsf{+3c}}}{\vphantom{\frac{1}{4}}B} = \underset{\color{red}{\mathsf{+3c}}}{\frac{1}{4}kc^2-3c} \\[16pt] \underset{\color{red}{\mathsf{\small{\times}\normalsize 4}}}{\vphantom{\frac{1}{4}}B+3c} = \underset{\color{red}{\mathsf{\small{\times}\normalsize 4}}}{\frac{1}{4}kc^2} \\[16pt] \underset{\color{red}{\mathsf{\div c^2}}}{4(B+3c)} = \underset{\color{red}{\mathsf{\div c^2}}}{\vphantom{4(B+3c)}kc^2} \\[10pt] \frac{4(B+3c)}{c^2} = k \\[6pt] k = \frac{4(B+3c)}{c^2} \end{gather} $$

Note 1: The \(\times 4\) and \(\div c^2\) steps may be done in either order.

Note 2: The final answer may also be given in this alternative form:

$$ \begin{gather} k = \frac{4B}{c^2}+\frac{12}{c} \end{gather} $$