Substitution and Rearranging Formulas
Theory
A formula links several quantities. To find one of them, substitute the values you know and solve, or change the subject so the letter you want is on its own. Use inverse operations in reverse order, doing the same to both sides, just as when solving an equation. This Year 10 maths page covers fractions and brackets, squares, cubes and roots, a subject in the denominator, and substituting into the new formula. Curriculum: VC2M10A05 / AC9M10A01.
The subject of a formula is the letter on its own on one side. In \(A=\pi r^2\), \(A\) is the subject.
To change the subject, treat the formula like an equation to solve for the new letter. Every step is done to both sides.
Use inverse operations: \(+\) and \(-\) undo each other, \(\times\) and \(\div\) undo each other, and squaring and square-rooting undo each other (as do cubing and cube-rooting).
Undo the operations on the new subject in reverse order: the last thing done to it is the first thing undone.
To substitute into a formula, replace each letter with its value. If the unknown is not the subject, either substitute first and solve the equation that is left, or change the subject first and then substitute. Both give the same answer.
| Operation on the subject | Undo it by |
|---|---|
| \(+a\) | \(-a\) |
| \(\times a\) | \(\div a\) |
| divided by \(a\) | \(\times a\) |
| squared | \(\pm\sqrt{\ }\) (or \(+\sqrt{\ }\) if it must be positive) |
| square root | square both sides |
| cubed | \(\sqrt[3]{\ }\) |
Subject in a reciprocal: combine the other side into one fraction, then flip both sides.
How to change the subject
- Clear fractions by multiplying both sides by the denominator, and square both sides to remove a square root containing the subject.
- Deal with brackets: divide both sides by the number in front of the bracket, or expand it first.
- Isolate the subject's term by moving every other term to the other side.
- Divide by the coefficient, then take a root if the subject is squared or cubed.
- Check by substituting simple numbers into the old and new formulas.
Using and checking a rearranged formula
- Substitute into the rearranged formula once, rather than rearranging with numbers each time (Examples 2 and 4).
- Square every factor: \((3k)^2 = 9k^2\), not \(3k^2\).
- Units: convert where needed after substituting, for example m/s to km/h by multiplying by \(3.6\).
- Spotting an error: test a rearrangement with numbers. If the values do not satisfy the original formula, find the faulty step. A square root cannot be taken term by term: \(\sqrt{a^2-b^2}\ne a-b\).
- Restrictions: a division by a bracket needs that bracket not to be zero; state conditions such as \(y\ne 2\) when asked.
Undo the operations on \(r\) in reverse order: \(\div 3\), then \(\times\pi h\), then squaring.
| \(3V\) | \(=\) | \(\pi r^2h\) |
| \(\dfrac{3V}{\pi h}\) | \(=\) | \(r^2\) |
| \(r\) | \(=\) | \(\boldsymbol{\sqrt{\dfrac{3V}{\pi h}}}\) |
Only the positive root is needed, because a radius is positive.
(i) Make \(h\) the subject.
(ii) With \(g=9.8\), find the height that gives a speed of \(14\) m/s.
(i) To undo a square root, square both sides.
| \(v^2\) | \(=\) | \(2gh\) |
| \(h\) | \(=\) | \(\boldsymbol{\dfrac{v^2}{2g}}\) |
(ii) Substitute into the new formula.
| \(h\) | \(=\) | \(\dfrac{14^2}{2\times 9.8} = \dfrac{196}{19.6}\) |
| \(\) | \(=\) | \(\mathbf{10}\text{ m}\) |
Check: \(\sqrt{2\times 9.8\times 10}=\sqrt{196}=14\) β
(i) Make \(d\) the subject of \(F = \dfrac{k}{d^2}\), given \(d > 0\).
(ii) Make \(x\) the subject of \(y = 5 - 2x^3\).
(i) \(d\) is in the denominator. Multiply both sides by \(d^2\) first, then divide by \(F\) and take the square root.
| \(Fd^2\) | \(=\) | \(k\) |
| \(d^2\) | \(=\) | \(\dfrac{k}{F}\) |
| \(d\) | \(=\) | \(\boldsymbol{\sqrt{\dfrac{k}{F}}}\) |
Only the positive root is kept, because \(d > 0\).
(ii) Get the \(x^3\) term on its own, divide by its coefficient, then take the cube root.
| \(2x^3\) | \(=\) | \(5 - y\) |
| \(x^3\) | \(=\) | \(\dfrac{5 - y}{2}\) |
| \(x\) | \(=\) | \(\boldsymbol{\sqrt[3]{\dfrac{5 - y}{2}}}\) |
A cube root has only one value, so there is no \(\pm\). Check with \(x = 1\): \(y = 5 - 2 = 3\), and \(\sqrt[3]{\dfrac{5 - 3}{2}} = \sqrt[3]{1} = 1\) β
(i) Make \(r\) the subject.
(ii) \(\$5000\) grows to \(\$5600\) in \(3\) years. Find the interest rate.
(i) Divide by \(P\) to remove the bracket's multiplier, then undo the \(+1\) and the \(\times n\).
| \(\dfrac{A}{P}\) | \(=\) | \(1+rn\) |
| \(\dfrac{A}{P}-1\) | \(=\) | \(rn\) |
| \(r\) | \(=\) | \(\dfrac{A-P}{Pn}\) |
\(\dfrac{A}{P}-1=\dfrac{A-P}{P}\), so dividing by \(n\) gives \(\boldsymbol{r=\dfrac{A-P}{Pn}}\).
(ii)
| \(r\) | \(=\) | \(\dfrac{5600-5000}{5000\times 3} = \dfrac{600}{15\,000}\) |
| \(\) | \(=\) | \(0.04 = \mathbf{4\%}\text{ a year}\) |
Common pitfalls
Frequently asked questions
How do you change the subject of a formula?
Treat it like an equation and solve for the new letter. Undo the operations on that letter in reverse order, doing the same thing to both sides each time, until it is on its own.
How do you use a formula to find an unknown value?
Substitute every value you know, then solve the equation that is left for the unknown. If you need several answers from the same formula, change the subject first and then substitute each time.
How do you make a squared letter the subject?
Get the squared term on its own, then take the square root of both sides. Write \(\pm\), unless the letter stands for something that must be positive, such as a length.
How do you remove a square root when changing the subject?
Get the square root on its own on one side, then square both sides. Square every factor on the other side too.
How can I check my rearranged formula?
Choose simple values, work out the original subject with the old formula, then put the values into your new formula. You should get back the value you started with.