Using Formulas
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, and how to find a letter that is not on its own: substitute what you know, then solve the equation that is left.
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.
To find a letter that is not the subject, substitute first, then solve: undo the operations around the unknown in reverse order, doing the same to both sides.
Some formulas you will meet:
| Formula | Subject | What it gives |
|---|---|---|
| \(A = lb\) | \(A\) | area of a rectangle |
| \(P = 2(l + b)\) | \(P\) | perimeter of a rectangle |
| \(v = u + at\) | \(v\) | final speed |
| \(s = \dfrac{d}{t}\) | \(s\) | average speed |
When the unknown is not the subject, substitute first, then solve. For \(v = u + at\) with \(v = 20\), \(u = 8\) and \(t = 3\):
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.
Substitute \(P = 50\) and \(b = 9\). The unknown is inside the bracket, so undo the \(\times\, 2\) first: divide the whole of each side by \(2\).
| \(50\) | \(=\) | \(2(l + 9)\) |
| \(25\) | \(=\) | \(l + 9\) |
| \(l\) | \(=\) | \(25 - 9\) |
| \(\) | \(=\) | \(\boldsymbol{16}\) |
The length is 16 cm.
Check: \(2(16 + 9) = 2 \times 25 = 50\) \(\checkmark\)
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.
What if the unknown is inside a bracket?
Substitute the values you know, then undo the operations in reverse order. For \(50 = 2(l + 9)\), divide both sides by \(2\) to get \(25 = l + 9\), then subtract \(9\), so \(l = 16\).
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.