NSW HSC Mathematics Standard

Formulas and pronumerals

A formula is a rule that links quantities, written with letters called pronumerals. The formula $A = \tfrac{1}{2}h(a + b)$ gives the area of a trapezium, where $a$ and $b$ are the parallel sides and $h$ is the distance between them. Each letter stands for a number that changes from one case to the next, and the formula fixes how the quantities depend on each other.

Units matter as much as numbers. If $h$ is in centimetres and $a$ and $b$ are in metres, the formula gives nonsense, so every quantity has to be in a compatible unit before it is used.

Substituting into a formula

To substitute, replace each pronumeral with its value and then evaluate, following the order of operations: brackets first, then powers, then multiplication and division, then addition and subtraction.

For the trapezium with $a = 5$, $b = 9$ and $h = 6$:

The same steps work for a non-linear formula. A ball dropped from rest falls a distance $s = 4.9t^2$ metres in $t$ seconds, so after $t = 3$ seconds:

The power applies to the 3 alone, so $3^2$ is worked out before the multiplication by 4.9.

A negative value goes into the formula inside brackets. To convert using $C = \tfrac{5}{9}(F - 32)$ when $F = 14$:

A scientific calculator does this directly if the whole expression is typed with brackets in the right places, and the fraction button keeps $\tfrac{5}{9}$ exact. A common error is to type $5 \div 9 \times 14 - 32$, which ignores the brackets.

Finding a value that is not the subject

Often the formula is given with one pronumeral on its own and the value wanted is one of the others. Substitute the known values and solve the resulting equation.

The formula $C = \tfrac{5}{9}(F - 32)$ with $C = 20$ gives $20 = \tfrac{5}{9}(F - 32)$. Multiplying both sides by $\tfrac{9}{5}$ gives $F - 32 = 36$, so $F = 68$.

Changing the subject of a formula

The subject of a formula is the pronumeral on its own on one side. To change the subject, undo the operations applied to the required pronumeral in the reverse order to that in which they were applied, doing the same thing to both sides each time.

Linear formulas

Make $w$ the subject of $P = 2l + 2w$.

Make $a$ the subject of $v = u + at$.

When the wanted letter is inside brackets, either divide first or expand first. For $C = \tfrac{5}{9}(F - 32)$, multiply both sides by $\tfrac{9}{5}$ to get $F - 32 = \tfrac{9}{5}C$, then add 32, giving $F = \tfrac{9}{5}C + 32$.

Formulas of the form $y = ax^2 + c$

Non-linear rearrangement is limited to quadratics with no $x$ term. Subtract $c$, divide by $a$, and take the square root:

There are two square roots, and the context decides which one is used. For a length, a time or a number of items the negative root is rejected.

A ball falls from a tower and $s = 4.9t^2$. It takes how long to fall 78.4 metres? Here $c = 0$, so $t^2 = 78.4 \div 4.9 = 16$, and $t = 4$ seconds (the root $-4$ has no meaning for a time).

A second example has a constant: the cost of a circular rug is $C = 30r^2 + 25$ dollars, where $r$ is the radius in metres. If the rug costs 145 dollars then $30r^2 = 120$, $r^2 = 4$ and $r = 2$ metres.

The area formula $A = \pi r^2$ is rearranged the same way. With $A = 50\ \mathrm{cm^2}$, $r = \sqrt{50 \div \pi} = 3.99$ cm to two decimal places.

Turning a word problem into an equation

To solve a problem given in words, name the unknown, write the relationship as an equation, solve it, and then read the answer back in the context of the question.

A hall hire costs 120 dollars plus 35 dollars for each hour. The total for an event was 330 dollars. Let $t$ be the number of hours.

The hall was hired for 6 hours. Check by substitution: $120 + 35 \times 6 = 120 + 210 = 330$.

The solution may need interpreting. A party budget of 500 dollars covers a fixed 80 dollars plus 28 dollars a guest, so $80 + 28n = 500$ gives $28n = 420$ and $n = 15$ guests exactly. With a budget of 450 dollars, $28n = 370$ and $n = 13.2$. Guests are whole people, so at most 13 can be invited, and the cost for 13 guests is $80 + 28 \times 13 = 444$ dollars, which is within the budget.

Equations with the unknown on both sides collect the unknowns on one side first. If two plans have the cost $12 + 4x$ and $6x$, they cost the same when $12 + 4x = 6x$, so $12 = 2x$ and $x = 6$.

Using a spreadsheet with a formula

A spreadsheet applies a formula to many values at once. Each value sits in a cell named by its column and row, and a formula starts with an equals sign. If the side length of a square is in column A, the area is found in cell B2 by entering =A2^2. Copying the formula down the column changes the cell reference in each row, so =A3^2 appears in B3, and the whole column of areas is found without retyping.

A cell that has to stay fixed when a formula is copied is written with dollar signs in the spreadsheet, as in =A2*$D$1, where D1 holds a constant such as a price or a rate.