Using Formulae
Formulas such as \(v = u + at\) and \(A = \frac{1}{2}bh\) connect quantities in science, measurement and money. This NSW Year 9 Core maths subtopic substitutes values into formulas and solves the equations that result, under NESA outcome MA5‑EQU‑C‑01.
You will substitute values and evaluate, keeping the order of operations and taking care with negatives, as in \(s = ut + \frac{1}{2}at^2\) with \(u = -4\). You will also substitute every value except one and solve the equation that is left, finding a triangle's height from its area, a chicken's mass from its cooking time, a temperature in Fahrenheit from Celsius, and the height of a cylinder from its surface area. Other examples include a bakery's profit, a draining tank and a car's speed in metres per second. It is part of Year 9 Core Equations.
Theory
A formula is an equation that links quantities, such as \(A = lb\). This Year 9 maths page shows how to substitute values into a formula, how to solve for a letter that is not on its own, and how to change the subject of a formula.
A formula is a rule that links two or more quantities. In \(A = lb\), the area \(A\) of a rectangle is found from its length \(l\) and breadth \(b\).
The subject of a formula is the letter on its own, usually on the left. In \(A = lb\), the subject is \(A\).
To substitute, replace each letter by its value. Put a negative value in brackets. Then work it out using the order of operations.
If the unknown is not the subject, substitute everything you know. This gives an ordinary equation to solve.
Changing the subject means rearranging the formula so a different letter is on its own. This comes up if you are heading for the Advanced path.
Some formulas you will meet:
| Formula | Subject | Rearranged for another letter |
|---|---|---|
| \(A = lb\) | \(A\) | \(l = \dfrac{A}{b}\) |
| \(P = 2(l + b)\) | \(P\) | \(l = \dfrac{P}{2} - b\) |
| \(v = u + at\) | \(v\) | \(t = \dfrac{v - u}{a}\) |
| \(s = \dfrac{d}{t}\) | \(s\) | \(d = st\) |
Rearranging uses the same undoing steps as solving:
How to use a formula
- Write the formula and list the values you know, in matching units.
- Substitute each value, with negatives in brackets.
- Solve: if the unknown is the subject, just work it out. If not, solve the equation that is left.
- Answer with units, rounded as asked. Check by putting the answer back in.
(i) The area of a triangle is \(A = \dfrac{1}{2}bh\). Find \(A\) when \(b = 12\) and \(h = 7\).
(ii) Using \(K = \dfrac{1}{2}mv^2\), find \(K\) when \(m = 4\) and \(v = -3\).
The unknown is the subject, so substitute and work it out.
(i)
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 12 \times 7\) |
| \(\) | \(=\) | \(\boldsymbol{42}\) |
(ii) Put the negative in brackets. Square before multiplying.
| \(K\) | \(=\) | \(\dfrac{1}{2} \times 4 \times (-3)^2\) |
| \(\) | \(=\) | \(\dfrac{1}{2} \times 4 \times 9\) |
| \(\) | \(=\) | \(\boldsymbol{18}\) |
\((-3)^2 = 9\), not \(-9\).
Substitute \(C = 255\). This gives an equation to solve for \(h\).
| \(255\) | \(=\) | \(60h + 90\) |
| \(165\) | \(=\) | \(60h\) |
| \(h\) | \(=\) | \(\dfrac{165}{60}\) |
| \(\) | \(=\) | \(2.75 \text{ hours}\) |
The \(0.75\) is a part of an hour: \(0.75 \times 60 = 45\) minutes.
The job took 2 hours 45 minutes.
Substitute \(C = 50\), then divide both sides by \(2\pi\).
| \(50\) | \(=\) | \(2\pi r\) |
| \(r\) | \(=\) | \(\dfrac{50}{2\pi}\) |
| \(\) | \(=\) | \(7.9577\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.96} \textbf{ cm}\) |
On a calculator, type \(50 \div (2 \times \pi)\), with the brackets.
(i) Make \(l\) the subject of \(P = 2(l + b)\).
(ii) Use it to find \(l\) when \(P = 50\) and \(b = 9\).
(i) Undo the steps in reverse order, with letters instead of numbers. Divide by \(2\) first, then take off \(b\).
| \(P\) | \(=\) | \(2(l + b)\) |
| \(\dfrac{P}{2}\) | \(=\) | \(l + b\) |
| \(l\) | \(=\) | \(\boldsymbol{\dfrac{P}{2} - b}\) |
(ii)
| \(l\) | \(=\) | \(\dfrac{50}{2} - 9\) |
| \(\) | \(=\) | \(25 - 9\) |
| \(\) | \(=\) | \(\boldsymbol{16}\) |
Check: \(2(16 + 9) = 50\) \(\checkmark\)
Common pitfalls
Frequently asked questions
How do you substitute into a formula?
Replace each letter by its value, putting negative values in brackets. Then work out the answer using the order of operations.
What is the subject of a formula?
It is the letter on its own on one side of the formula. In \(A = lb\), the subject is \(A\).
How do you find a value that is not the subject?
Substitute all the values you know. You get an equation with one unknown. Solve it the usual way.
How do you change the subject of a formula?
Undo the operations around the letter you want, in reverse order, doing the same to both sides. For \(v = u + at\), subtract \(u\), then divide by \(a\): \(t = \dfrac{v - u}{a}\).
How do you convert hours to minutes in a formula answer?
Multiply the decimal part of the hours by \(60\). For \(2.75\) hours, \(0.75 \times 60 = 45\), so it is \(2\) hours \(45\) minutes.