Expressing one quantity as a function of others

What the question is asking

A question of this type describes a situation and asks for one quantity as a function of another: "express the area as a function of the width", "write the profit in terms of the number of items sold". The answer is a formula with exactly one variable in it, the one named, plus fixed numbers. If your formula still contains a second letter, you are not finished.

A four-step method

1. Name the variables. Write down what each letter means, with units: "$x$ = width in metres".
2. Write the target quantity in whatever variables are natural: area $= x \cdot y$, profit $=$ revenue $-$ cost.
3. Use the constraint the problem gives (a fixed perimeter, a fixed total, a geometric relation) to write the extra variable in terms of the chosen one, and substitute.
4. State the domain. Lengths are positive, counts are whole numbers, a square root needs a non-negative argument, and a denominator cannot be zero. The domain is part of the answer, and it often decides which option is right.

Common patterns

Situation	Typical formula
Fixed charge plus a rate	$C(x) = F + rx$ (a straight line)
Price falls as quantity rises	revenue $R(x) = x \cdot p(x)$, often a quadratic
Profit	$P(x) = R(x) - C(x)$
Rectangle with fixed perimeter $2s$	width $x$, length $s - x$, area $x(s - x)$
Rectangle with fixed area $S$	width $x$, length $\frac{S}{x}$, perimeter $2x + \frac{2S}{x}$
Right triangle with hypotenuse $h$	legs $x$ and $\sqrt{h^2 - x^2}$
Time for a distance at a speed	$t = \frac{d}{v}$

Two quantities are directly proportional when one is a fixed multiple of the other ($y = kx$), and inversely proportional when their product is fixed ($y = \frac{k}{x}$). Doubling a rectangle's width at fixed area halves its length: inverse proportion.

Units

Check that every term has the same unit before adding. A taxi fare with a per-kilometre rate cannot have the distance in metres, and a speed in kilometres per hour needs a time in hours: 20 minutes is $\frac{1}{3}$ hour, not $0.2$.

Rearranging a formula

Making another variable the subject is the same skill run backwards. Treat the other letters as fixed numbers and undo the operations in reverse order. When the variable you want appears in more than one place, collect all its terms on one side and factor it out:

$$y = \frac{3x + 2}{x - 1} \;\Rightarrow\; y(x - 1) = 3x + 2 \;\Rightarrow\; xy - 3x = y + 2 \;\Rightarrow\; x(y - 3) = y + 2 \;\Rightarrow\; x = \frac{y + 2}{y - 3}.$$

The new formula has its own restriction ($y \ne 3$), because you divided by $y - 3$. Check the rearrangement with one number: $x = 2$ gives $y = 8$, and $\frac{8 + 2}{8 - 3} = 2$.

Why the check matters

Pick one simple value of the variable, compute the quantity directly from the story, and compare it with your formula. A wrong formula often differs from the right one by a factor of 2, a sign, or a missing term, and one substitution tells them apart.