If $y$ is directly proportional to $x$, written $y \propto x$, introduce a constant $k$ to turn this into a working equation: $y = kx$. The method for any proportion problem is always the same four steps:

1. Write the proportionality statement, then convert it to an equation with $k$.
2. Substitute a known pair of values to solve for $k$.
3. Rewrite the equation using that value of $k$.
4. Substitute the new value to answer the question.

Example: Pay ($P$) is proportional to hours worked ($H$). If $P=30$ when $H=3$: $30 = 3k \Rightarrow k=10$, so $P = 10H$. Pay for 100 hours: $P = 1000$.

If $y$ is inversely proportional to $x$ ($y \propto \frac1x$), increasing $x$ decreases $y$ by the same factor. The equation is $y = \dfrac{k}{x}$, and the product $xy$ is always constant and equal to $k$.

Example (work-rate): Time $t$ to paint a house is inversely proportional to the number of painters $P$. If $t=8$ hours with $P=4$: $8 = \dfrac{k}{4} \Rightarrow k=32$. So $t = \dfrac{32}{P}$, and with 8 painters, $t=4$ hours. Here $k=32$ represents the total "person-hours" of work needed — useful when painters join or leave partway through a job.

The same method extends to $y \propto x^n$ for any power, including roots:

Example: If $T=32$ when $U=8$ and $T=kU^2$: $32 = k(8)^2 = 64k \Rightarrow k = 0.5$.

If $T = k\sqrt{L}$ and $L$ increases by 40%, the new length is $L_{new}=1.4L$, so:

Since $\sqrt{1.4}\approx1.1832$, $T$ increases by about $18.3\%$. The scale factor from the $x$-change passes through the power or root in the formula — it does not apply directly to $y$ unless the relationship is linear.

• $y=kx$: straight line through the origin, gradient $k$.
• $y=kx^2$: a parabola through the origin, but curving away faster.
• $y=k/x$: a hyperbola with two branches that approach (but never touch) both axes — $x\neq0$ and $y\neq0$ always.

Worked Example

The time period $T$ of a pendulum is directly proportional to $\sqrt{L}$, its length. A pendulum of length $2.25\text{ m}$ has a period of $3\text{ s}$. Find $T$ for a pendulum of length $9\text{ m}$.

1. $T = k\sqrt{L}$, so $3 = k\sqrt{2.25} = 1.5k \Rightarrow k = 2$.
2. Equation: $T = 2\sqrt{L}$.
3. Substitute $L=9$: $T = 2\sqrt{9} = 2\times3 = 6\text{ s}$.