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 \(m\).

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

Video solution by Clelland Maths 

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 \(c^2\).

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

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.

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

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!

Video solution by Clelland Maths 

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

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

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.

$$ \begin{eqnarray} a = b \sqrt{c}-d \\[6pt] a+d = b \sqrt{c} \\[6pt] \small\frac{a+d}{b}\normalsize = \sqrt{c} \\[6pt] \small\left(\frac{a+d}{b}\right)^2\normalsize = c \\[6pt] c = \small\left(\frac{a+d}{b}\right)^2\normalsize \end{eqnarray} $$

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

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

Video solution by Clelland Maths 

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:

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

Video solution by Clelland Maths 

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

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

Video solution by Clelland Maths 

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

Video solution by Clelland Maths 

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

Video solution by Clelland Maths 

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

Video solution by Clelland Maths 

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

Video solution by Clelland Maths 

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.

Video solution by Clelland Maths 

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

Video solution by Clelland Maths 

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

Video solution by Clelland Maths 

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

Video solution by Clelland Maths 

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

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

Video solution by Clelland Maths 

$$ \begin{eqnarray} B = \small\frac14\normalsize kc^2-3c\\[6pt] B+3c = \small\frac14\normalsize kc^2\\[6pt] 4(B+3c) = kc^2\\[6pt] \small\frac{4(B+3c)}{c^2}\normalsize = k \\[6pt] k = \small\frac{4(B+3c)}{c^2}\normalsize \end{eqnarray} $$