Literal Equations
A literal equation is a formula with several pronumerals, such as \(v=u+at\). Changing the subject rearranges it so that a different pronumeral stands alone, using the same inverse operations as solving an equation. In NSW Year 10 Advanced maths the formulas include squares, roots, fractions and the subject appearing twice (NESA outcome MA5‑EQU‑P‑02).
You will change the subject of formulas such as \(A=\pi r^2\), \(P=2(l+w)\), \(V=\dfrac{1}{3}Ah\) and \(v=\sqrt{u^2+2as}\). You will collect terms when the subject appears twice, as in \(p=\dfrac{2q-1}{q+3}\), and complete the square when the subject is in a squared term and a plain term. Practical formulas include a ladder against a wall, compound interest, body mass index and resistors in parallel. It is part of Year 10 Advanced Non Linear Relationships Functions and Graphs.
Every question with a fully worked solution.
- Literal Equations - Video - Literal Equations (Solving for a Variable) Watch
Theory
A literal equation is a formula with several pronumerals. To change the subject, undo the operations around the new subject with their opposites, in reverse order, keeping both sides balanced. This Year 10 Advanced maths page covers powers and roots, the subject in more than one term, reciprocals, completing the square and substituting values afterwards.
The subject of a formula is the pronumeral on its own on one side. In \(A = \pi r^2\), \(A\) is the subject. A literal equation is an equation in letters, and solving it for one letter is called changing the subject or transposing the formula.
The rules are the same as for solving a number equation: do the same thing to both sides. The opposite of adding is subtracting, of multiplying is dividing, of squaring is taking the square root, and of cubing is taking the cube root.
If the new subject appears in more than one term, collect those terms on one side and take the subject out as a common factor. If it is in a denominator, multiply both sides by that denominator first.
Taking a square root gives two signs, \(\pm\). Keep only the sign that fits the condition given, such as \(r \gt 0\) for a length.
It is used in every topic of Advanced maths in Years 11 and 12.
Opposite operations:
| In the formula | Undo it by | Example |
|---|---|---|
| \(+\,c\) | \(-\,c\) | \(P = Q + 7 \Rightarrow Q = P - 7\) |
| \(\times\, c\) | \(\div\, c\) | \(F = ma \Rightarrow a = \dfrac{F}{m}\) |
| \(x^2\) | \(\pm\sqrt{\ }\) | \(A = 6s^2 \Rightarrow s = \sqrt{\dfrac{A}{6}}\ (s \gt 0)\) |
| \(\sqrt{\ }\) | square both sides | \(v = \sqrt{2gh} \Rightarrow h = \dfrac{v^2}{2g}\) |
| \(x^3\) | \(\sqrt[3]{\ }\) | \(V = 2x^3 \Rightarrow x = \sqrt[3]{\dfrac{V}{2}}\) |
| \(\dfrac{1}{x}\) | flip both sides | \(\dfrac{1}{u} = \dfrac{p}{q} \Rightarrow u = \dfrac{q}{p}\) |
How to change the subject
- Circle the new subject. Note any condition, such as \(h \gt 0\).
- Clear fractions if the subject is in a denominator: multiply both sides by it.
- Expand brackets that contain the subject.
- Collect every term with the subject on one side and the rest on the other.
- Factorise out the subject if it is in two or more terms, then divide by the bracket.
- Undo powers and roots last: square to remove a square root, take a root to remove a power, and choose the sign.
- Check by picking simple numbers, or by substituting back.
Special cases
- Several fractions: multiply every term by the lowest common denominator, or put the other side over one denominator before flipping (Example 3).
- Square root over the whole formula, such as \(v = \sqrt{u^2 + 2as}\): square both sides first, then \(2as = v^2 - u^2\) and \(s = \dfrac{v^2 - u^2}{2a}\). Never split the root into \(u + \sqrt{2as}\).
- Subject in a squared term and a plain term: complete the square so the subject appears once (Example 4).
- Rearrange, then substitute: when a question asks for a value, change the subject first, then substitute, and round only at the end.
(i) The area of a trapezium is \(A = \dfrac{h(a+b)}{2}\). Make \(b\) the subject.
(ii) The time for one swing of a pendulum is \(T = 2\pi\sqrt{\dfrac{L}{g}}\). Make \(L\) the subject, then find \(L\) when \(T = 2.4\) and \(g = 9.8\), correct to two decimal places.
(i) \(b\) was added to \(a\), multiplied by \(h\), then divided by \(2\). Undo those steps from the last one back.
| \(2A\) | \(=\) | \(h(a+b)\) |
| \(\dfrac{2A}{h}\) | \(=\) | \(a+b\) |
| \(b\) | \(=\) | \(\boldsymbol{\dfrac{2A}{h}-a}\) |
(ii) Divide by \(2\pi\), square both sides to remove the root, then multiply by \(g\).
| \(\dfrac{T}{2\pi}\) | \(=\) | \(\sqrt{\dfrac{L}{g}}\) |
| \(\dfrac{T^2}{4\pi^2}\) | \(=\) | \(\dfrac{L}{g}\) |
| \(L\) | \(=\) | \(\boldsymbol{\dfrac{gT^2}{4\pi^2}}\) |
| \(L\) | \(=\) | \(\dfrac{9.8\times 2.4^2}{4\pi^2}\) |
| \(\) | \(=\) | \(1.4298\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1.43}\) |
Squaring \(\dfrac{T}{2\pi}\) squares the \(2\) and the \(\pi\) as well: \((2\pi)^2 = 4\pi^2\). The pendulum is about \(1.43\) m long.
(i) Make \(m\) the subject of \(km + 3 = 2m - k\).
(ii) Make \(t\) the subject of \(w = \dfrac{3t-2}{t+4}\).
(i) Gather every term containing \(m\) on one side and everything else on the other. Then take out \(m\) as a common factor.
| \(km - 2m\) | \(=\) | \(-k - 3\) |
| \(m(k-2)\) | \(=\) | \(-(k+3)\) |
| \(m\) | \(=\) | \(\dfrac{-(k+3)}{k-2}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{k+3}{2-k}}\) |
Multiplying the top and bottom by \(-1\) removes the minus signs. Both forms are correct.
(ii) \(t\) is in the numerator and the denominator. Multiply by the denominator first, then expand.
| \(w(t+4)\) | \(=\) | \(3t-2\) |
| \(wt + 4w\) | \(=\) | \(3t-2\) |
| \(wt - 3t\) | \(=\) | \(-2 - 4w\) |
| \(t(w-3)\) | \(=\) | \(-(4w+2)\) |
| \(t\) | \(=\) | \(\boldsymbol{\dfrac{4w+2}{3-w}}\) |
Check with \(t = 2\): \(w = \dfrac{4}{6} = \dfrac{2}{3}\), and \(\left(4\times\dfrac{2}{3}+2\right) \div \left(3-\dfrac{2}{3}\right) = \dfrac{14}{3}\div\dfrac{7}{3} = 2\).
Get the \(y\) term alone, combine the other side into one fraction, then flip both sides.
| \(\dfrac{1}{y}\) | \(=\) | \(\dfrac{2}{x} - \dfrac{3}{z}\) |
| \(\dfrac{1}{y}\) | \(=\) | \(\dfrac{2z - 3x}{xz}\) |
| \(y\) | \(=\) | \(\boldsymbol{\dfrac{xz}{2z-3x}}\) |
| \(y\) | \(=\) | \(\dfrac{4\times 10}{2\times 10 - 3\times 4}\) |
| \(\) | \(=\) | \(\dfrac{40}{8} = \mathbf{5}\) |
Check: \(\dfrac{2}{4} - \dfrac{1}{5} = 0.5 - 0.2 = 0.3 = \dfrac{3}{10}\). Writing \(y = \dfrac{x}{2} - \dfrac{z}{3}\) would be wrong: you cannot flip the terms one at a time.
\(x\) appears in an \(x^2\) term and an \(x\) term. Complete the square so that \(x\) appears only once. Half of \(-6\) is \(-3\), and \((-3)^2 = 9\).
| \(y\) | \(=\) | \(x^2 - 6x + 9 - 9 + 2\) |
| \(y\) | \(=\) | \((x-3)^2 - 7\) |
| \((x-3)^2\) | \(=\) | \(y + 7\) |
| \(x - 3\) | \(=\) | \(\pm\sqrt{y+7}\) |
Because \(x \ge 3\), \(x - 3\) is never negative, so keep the positive root.
| \(x\) | \(=\) | \(\boldsymbol{3 + \sqrt{y+7}}\) |
| \(x\) | \(=\) | \(3 + \sqrt{9+7}\) |
| \(\) | \(=\) | \(3 + 4 = \mathbf{7}\) |
Check: \(7^2 - 6\times 7 + 2 = 49 - 42 + 2 = 9\). Without the condition \(x \ge 3\) the answer would be \(x = 3 \pm \sqrt{y+7}\).
Common pitfalls
Frequently asked questions
What does it mean to change the subject of a formula?
It means rearranging the formula so that a different pronumeral is on its own on one side. For example, \(A = lw\) becomes \(w = \dfrac{A}{l}\).
How do you rearrange a formula?
Work out the operations that were done to the new subject and undo them with their opposites, starting with the last one. Do every step to both sides of the formula.
What do you do if the subject appears twice?
Move every term containing the subject to one side and all other terms to the other side. Then take the subject out as a common factor and divide by what is left in the bracket.
How do you make a pronumeral the subject when it is in the denominator?
Multiply both sides by the denominator to bring the pronumeral to the top. Then expand, collect the terms with the subject and factorise if needed.
Should I write plus or minus when I take a square root?
Taking a square root always gives two possible signs. Keep plus or minus unless the question says the subject is positive or negative, such as a length, which must be positive.