Using And Transposing Formulas
Using and transposing formulas is a foundational skill assumed for Queensland Year 11 Mathematical Methods (QCAA). Using a formula means substituting known values to find an unknown, while transposing means rearranging it to make a different variable the subject.
You will substitute carefully and change the subject of linear, squared and fractional formulas, working with everyday relationships from science and finance to build lasting algebraic fluency.
Theory
In Year 11 Mathematical Methods (QCAA), using a formula means substituting known values to evaluate an unknown, and transposing (or changing the subject of) a formula means rearranging it to make a different variable the subject. This page reviews substituting into formulas and transposing linear, squared and fractional formulas such as \(v=u+at\), \(A=6s^2\) and \(v^2=u^2+2as\).
A formula is an equation linking several quantities, such as \(A=\dfrac{1}{2}bh\). The variable on its own is the subject. To use a formula, substitute the known values and evaluate.
To transpose a formula (change its subject) we rearrange it with inverse operations until the required variable stands alone — exactly the technique used to solve a linear equation, but keeping the other letters as pronumerals.
Some formulas involve a square or a fraction. To free a squared variable, take the square root of both sides; to free a variable in a denominator, multiply both sides by that denominator.
Transposing a two-step linear formula, making \(a\) the subject of \(v=u+at\):
Freeing a squared variable by taking the positive square root (a length is positive):
Clearing a denominator, making \(m\) the subject of the density formula:
How to transpose a formula
- Locate the variable you want as the new subject and the operations attached to it.
- Undo those operations in reverse order, doing the same to both sides (clear fractions first if the variable is in a denominator).
- Root if needed: take the (positive) square root once the squared term is alone.
- Substitute any given values into the rearranged formula to find the number.
Substitute \(b=12\) and \(h=5\):
| \(A\) | \(=\) | \(\dfrac12(12)(5)\) |
| \(=\) | \(6\times5\) | |
| \(=\) | \(30\) |
Area \(=30\ \text{cm}^2\).
Subtract \(u\) from both sides:
| \(v-u\) | \(=\) | \(at\) |
Divide both sides by \(t\):
| \(\dfrac{v-u}{t}\) | \(=\) | \(a\) |
Reading it the usual way round gives the new subject.
\(a=\dfrac{v-u}{t}\).
Divide both sides by \(6\):
| \(\dfrac{A}{6}\) | \(=\) | \(s^2\) |
Take the positive square root (a side length is positive):
| \(s\) | \(=\) | \(\sqrt{\dfrac{A}{6}}\) |
Substitute \(A=54\):
| \(s\) | \(=\) | \(\sqrt{\dfrac{54}{6}}\) |
| \(=\) | \(\sqrt{9}\) | |
| \(=\) | \(3\) |
\(s=\sqrt{\dfrac{A}{6}}\); when \(A=54\), \(s=3\ \text{cm}\).
Subtract \(u^2\) from both sides:
| \(v^2-u^2\) | \(=\) | \(2as\) |
Divide both sides by \(2a\):
| \(\dfrac{v^2-u^2}{2a}\) | \(=\) | \(s\) |
Substitute \(u=4,\ v=10,\ a=3\):
| \(s\) | \(=\) | \(\dfrac{10^2-4^2}{2(3)}\) |
| \(=\) | \(\dfrac{100-16}{6}\) | |
| \(=\) | \(\dfrac{84}{6}\) | |
| \(=\) | \(14\) |
\(s=\dfrac{v^2-u^2}{2a}\); when \(u=4,\,v=10,\,a=3\), \(s=14\ \text{m}\).
Common pitfalls
Frequently asked questions
What does it mean to transpose a formula?
To transpose (or change the subject of) a formula is to rearrange it so a chosen variable stands alone on one side, using inverse operations.
How do you make a variable the subject of a formula?
Undo the operations attached to it in reverse order, doing the same to both sides. For \(v=u+at\), subtract \(u\) then divide by \(t\) to get \(a=\dfrac{v-u}{t}\).
How do you transpose a formula with a square in it?
Isolate the squared term, then take the square root of both sides. For \(A=6s^2\), divide by \(6\) first, giving \(s=\sqrt{\dfrac{A}{6}}\).
How do you transpose a formula with a fraction?
Multiply both sides by the denominator to clear it. For \(D=\dfrac{m}{V}\), multiply by \(V\) to get \(m=DV\).
Do you keep the plus-or-minus when taking a square root?
In these physical formulas the variable (a length, a side) is positive, so we take the positive root only, such as \(s=3\) cm.
What is the difference between using and transposing a formula?
Using a formula substitutes known values to evaluate the subject; transposing rearranges the formula first so a different variable becomes the subject.