Using Formulae
A formula is an equation that links quantities, so substituting the known values leaves an equation to solve for the missing one. This NSW Year 9 Advanced maths subtopic uses formulas from science, geometry and everyday life (NESA outcome MA5‑EQU‑C‑01).
You will substitute into formulas such as \(s = ut + \frac{1}{2}at^2\) and \(v = u + at\), and solve for the unknown in area, temperature, cooking-time and profit formulas. Harder questions find a speed from kinetic energy, use the lens formula \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\), find a radius from a cylinder's surface area, and find when a Fahrenheit reading is twice the Celsius reading. It is part of Year 9 Advanced Equations.
Theory
A formula is an equation that links several quantities. This Year 9 Advanced maths page shows how to substitute the values you know and solve for the one you want, wherever it sits: on its own, inside a bracket, squared or in a fraction. It also covers converting units first and changing the subject of a formula.
A formula such as \(A = \dfrac{1}{2}bh\) is a rule connecting quantities. The pronumeral on its own on the left is the subject.
If you want the subject, substitute and evaluate. If you want one of the other quantities, substitute everything you know and solve the equation that is left, using inverse operations. Put brackets round substituted values, especially negatives: \(u = -4\) in \(ut\) is \((-4)t\).
To change the subject, treat the formula as an equation and undo the operations around the new subject, in reverse order, until it is on its own.
Units must match the formula. If \(T\) is in minutes, a time of \(2\) hours goes in as \(120\). Convert before you substitute.
If the unknown is squared, take the square root at the end: \(x^2 = 25\) gives \(x = \pm 5\). Reject the negative root when the quantity is a length, a speed or another quantity that cannot be negative.
Undoing operations to isolate a pronumeral:
| Formula | Rearranged |
|---|---|
| \(y = mx + c\) | \(x = \dfrac{y - c}{m}\) |
| \(P = 2(l + w)\) | \(w = \dfrac{P}{2} - l\) |
| \(V = \dfrac{1}{3}Ah\) | \(h = \dfrac{3V}{A}\) |
| \(A = \pi r^2\) | \(r = \sqrt{\dfrac{A}{\pi}}\) (\(r > 0\)) |
| \(d = st\) | \(s = \dfrac{d}{t}\), \(t = \dfrac{d}{s}\) |
How to use a formula
- List each letter with its value and unit, and mark the one you want.
- Convert units so they match the formula.
- Substitute with brackets round each value, and work out every number you can (powers first, then products).
- Solve the equation that is left with inverse operations. For a squared unknown, isolate the square, then take the square root and choose the sign that makes sense. For a reciprocal, isolate the fraction and flip both sides.
- Answer with units and the rounding asked for, and check by substituting back.
Harder variations
- Changing the subject: the same steps with letters instead of numbers.
- One quantity in terms of another ("the height is \(3\) times the radius"): replace it by its expression, such as \(h = 3r\), before solving.
- A condition between two readings ("\(F\) is twice \(C\)"): write both in terms of one pronumeral and form one equation.
Substitute the three known values, work out the numbers, then solve for \(a\).
| \(14^2\) | \(=\) | \(6^2 + 2 \times a \times 20\) |
| \(196\) | \(=\) | \(36 + 40a\) |
| \(160\) | \(=\) | \(40a\) |
| \(a\) | \(=\) | \(\mathbf{4}\) |
The acceleration is \(4\) m/s². Check: \(36 + 2 \times 4 \times 20 = 196 = 14^2\).
| \(447.50\) | \(=\) | \(85t + 150\) |
| \(297.50\) | \(=\) | \(85t\) |
| \(t\) | \(=\) | \(\dfrac{297.50}{85} = 3.5\) |
\(t\) is in hours. \(0.5\) hours is \(0.5 \times 60 = 30\) minutes, so the booking was \(\mathbf{3}\) hours \(\mathbf{30}\) minutes.
Check: \(85 \times 3.5 + 150\) \(= 297.50 + 150\) \(= 447.50\).
Substitute, make \(r^2\) the subject, then take the square root.
| \(50\) | \(=\) | \(\pi r^2\) |
| \(r^2\) | \(=\) | \(\dfrac{50}{\pi} = 15.915\ldots\) |
| \(r\) | \(=\) | \(\pm\sqrt{15.915\ldots}\) |
| \(\) | \(=\) | \(\pm 3.989\ldots\) |
A radius is a length, so it cannot be negative: \(r \approx \mathbf{4.0}\) cm.
Keep \(\dfrac{50}{\pi}\) in the calculator; rounding it to \(16\) first still gives \(4.0\) here, but early rounding often changes an answer.
(i) Make \(t\) the subject of \(v = u + at\).
(ii) Use your formula to find \(t\) when \(v = 25\), \(u = 7\) and \(a = 3.6\).
(i) Undo the operations on \(t\) in reverse order: subtract \(u\), then divide by \(a\).
| \(v - u\) | \(=\) | \(at\) |
| \(t\) | \(=\) | \(\dfrac{v - u}{a}\) |
(ii)
| \(t\) | \(=\) | \(\dfrac{25 - 7}{3.6}\) |
| \(\) | \(=\) | \(\dfrac{18}{3.6} = \mathbf{5}\) |
A rearranged formula is worth it when you need the same unknown many times.
Common pitfalls
Frequently asked questions
How do you substitute into a formula?
Replace each letter by its value, putting brackets round each number, especially negatives. Then follow the order of operations: brackets, powers, multiplication and division, then addition and subtraction.
How do you find a value that is not the subject of the formula?
Substitute all the values you know. The formula becomes an equation with one unknown. Solve it with inverse operations, then check by substituting back.
How do you change the subject of a formula?
Treat it like an equation. Undo the operations around the new subject in reverse order, doing the same to both sides each time, until the new subject is on its own on one side.
Why do you reject the negative square root?
Squaring hides the sign, so \(x^2 = 16\) has two solutions, \(4\) and \(-4\). If \(x\) is a length, time or speed, the negative value has no meaning, so only the positive root is the answer.
Do units matter when using a formula?
Yes. Every quantity must be in the unit the formula expects. Convert hours to minutes or kilometres to metres before substituting, and give the answer in the unit that is asked for.