Substitution and Rearranging Formulas
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. To find an unknown value, substitute what you know and solve. This Year 10A maths page (Victorian Curriculum 10A) covers powers and roots, the subject in more than one term, reciprocals, and substituting values to find an unknown. Curriculum: VC2M10A05 / AC9M10A01.
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.
To find an unknown that is not the subject, substitute the known values and solve the equation that is left, or change the subject first and then substitute.
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 \(u + \sqrt{2as}\).
- Substitute, then solve: when the unknown is not the subject, substitute the known values and solve for it (Example 4).
- Rearrange, then substitute: when you need the same unknown several times, 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 surface area of a sphere is \(A = 4\pi r^2\). Make \(r\) the subject, then find \(r\) when \(A = 200\), 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) \(r\) was squared, then multiplied by \(4\pi\). Divide by \(4\pi\) first, then take the square root. A radius is positive, so keep only the positive root.
| \(\dfrac{A}{4\pi}\) | \(=\) | \(r^2\) |
| \(r\) | \(=\) | \(\boldsymbol{\sqrt{\dfrac{A}{4\pi}}}\) |
| \(r\) | \(=\) | \(\sqrt{\dfrac{200}{4\pi}}\) |
| \(\) | \(=\) | \(3.9894\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{3.99}\) |
Round only at the end. Check: \(4\pi \times 3.9894^2 \approx 200\) β
(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.
(i) The distance travelled by an accelerating object is \(s = ut + \dfrac{1}{2}at^2\). Find \(a\) when \(s = 100\), \(u = 4\) and \(t = 5\).
(ii) The volume of a cylinder is \(V = \pi r^2 h\). Find the radius \(r\) when \(V = 500\) and \(h = 8\), correct to one decimal place.
(i) \(a\) is not the subject. Substitute the three known values, then solve the equation for \(a\).
| \(100\) | \(=\) | \(4 \times 5 + \dfrac{1}{2} \times a \times 5^2\) |
| \(100\) | \(=\) | \(20 + 12.5a\) |
| \(80\) | \(=\) | \(12.5a\) |
| \(a\) | \(=\) | \(\mathbf{6.4}\) |
Check: \(4 \times 5 + \dfrac{1}{2} \times 6.4 \times 25 = 20 + 80 = 100\) β
(ii) Substitute, get \(r^2\) on its own, then take the positive square root.
| \(500\) | \(=\) | \(\pi r^2 \times 8\) |
| \(r^2\) | \(=\) | \(\dfrac{500}{8\pi}\) |
| \(\) | \(=\) | \(19.89\ldots\) |
| \(r\) | \(=\) | \(\sqrt{19.89\ldots}\) |
| \(\) | \(\approx\) | \(\mathbf{4.5}\) |
Keep the full value of \(r^2\) in the calculator and round only the final answer.
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.