R13 - Forming proportion equations (Higher) | AQA GCSE Mathematics

Turn a proportion statement into an equation, use a known pair of values to find its constant, then solve the completed model. You will work with direct and inverse proportion, including powers and roots, and explain how changing one quantity affects another.

Constructing a direct proportion equation (Higher)

When $x=2$ , $y=24$ . Can you predict $y$ when $x=4$ ? That single pair is not enough. Direct proportion, proportion to a square and inverse proportion all give different answers while fitting the same starting pair.

You already know

The symbol $\propto$ means “is proportional to”. Direct proportion has the form $y=kx$ ; inverse proportion has the form $y=k/x$ , with $x\ne0$ . A5 covered rearranging equations to change the subject.

A proportion statement tells you the form of the equation. A known pair then fixes the constant $k$ . For example, “ $y$ is directly proportional to $x$ ” gives $y=kx$ , not $y=x$ . The multiplier is still unknown until you use the data.

Constructing a direct model

If $y\propto x$ , write

y=kx\qquad k=\frac{y}{x}\quad(x\ne0)

Use a known non-zero input and its matching output to find $k$ . Keep that same constant for all pairs in the model.

Not on the formulae sheet: learn it.

Write the completed equation before using a new value. This keeps the two stages of the calculation separate.

Worked example 1: finding a new output

Question. $y$ is directly proportional to $x$ . When $x=6$ , $y=18$ . Work out an equation connecting $y$ and $x$ , then work out $y$ when $x=10$ . Non-calculator.

Plan. Write $y=kx$ , substitute the known pair to find $k$ , then use the completed rule.

\begin{aligned} y&=kx &&\text{direct-proportion form}\\ 18&=6k &&\text{substitute the known pair}\\ k&=18\div6=3 &&\text{find the multiplier}\\ y&=3x &&\text{completed equation}\\ y&=3\times10=30 &&\text{use the new input} \end{aligned}

Why use $18$ and $6$ together? They are one matching pair. Combining $18$ with the new input $10$ would change the model.

Answer. $y=3x$ ; at $x=10$ , $y=30$ .

Check. The completed equation gives $3\times6=18$ for the original pair.

Change only which variable you are finding. The model and known pair stay the same.

Worked example 2: finding a new input

Question. $y$ is directly proportional to $x$ . When $x=6$ , $y=18$ . Work out $x$ when $y=42$ . Non-calculator.

Plan. Recover the same constant, then solve the completed equation for $x$ .

\begin{aligned} y&=kx &&\text{choose the model}\\ k&=18\div6=3 &&\text{use the known pair}\\ 42&=3x &&\text{substitute the new output}\\ x&=42\div3=14 &&\text{divide by the multiplier} \end{aligned}

Why divide at the last step? The new output is three times the unknown input. Division undoes that multiplication.

Answer. $x=14$ .

Check. $3\times14=42$ .

A constant may have units in a physical model. If $C=kh$ gives pounds from hours, $k$ is a rate in pounds per hour. Use consistent units before finding it.

Watch out

Wrong: $y\propto x$ , so $y=x$ . Right: insert $k$ , then find it from a matching pair. Proportion specifies a relationship, not necessarily equality.

Find the constant before solving for a new input.

Direct proportion to powers and roots (Higher)

“Proportional to the square of $x$ ” means the multiplier acts on $x^2$ , not on $x$ . Read the entire phrase before writing the equation.

If $y=3x^2$ , doubling $x$ multiplies $y$ by $2^2=4$ . If $y=3x^3$ , doubling $x$ multiplies $y$ by $2^3=8$ . A power changes the way the quantities scale, even though the constant multiplier stays fixed.

Direct proportion to a specified power or root

\begin{array}{c|c} \text{Statement}&\text{Equation}\\\hline y\propto x^2&y=kx^2\\ y\propto x^3&y=kx^3\\ y\propto\sqrt{x}&y=k\sqrt{x}\\ y\propto\sqrt[3]{x}&y=k\sqrt[3]{x} \end{array}

Evaluate the specified power or root before dividing to find $k$ . For real square roots, $x\geqslant0$ ; a non-zero known root is needed to determine $k$ by division.

Not on the formulae sheet: learn it.

The input itself is not always the quantity directly proportional to the output. In the first example it is the square of the input.

Worked example 3: proportion to a square

Question. $y$ is directly proportional to $x^2$ . When $x=4$ , $y=12$ . Work out $y$ when $x=16$ . Non-calculator.

Plan. Write $y=kx^2$ , find $k$ using $4^2$ , then use $16^2$ .

\begin{aligned} y&=kx^2 &&\text{translate the statement}\\ 12&=k\times4^2=16k &&\text{square the known input}\\ k&=\frac{12}{16}=\frac34 &&\text{find the constant}\\ y&=\frac34\times16^2 &&\text{use the completed equation}\\ &=\frac34\times256=192 &&\text{evaluate} \end{aligned}

Why divide by $16$ , not $4$ , to find $k$ ? The output is proportional to the square, so the known multiplying quantity is $4^2$ .

Answer. $y=192$ .

Check. The input multiplied by $4$ ; the output multiplied by $4^2=16$ . Indeed, $12\times16=192$ .

Change only the relationship from a square to a square root. Keep both the known pair and the requested input unchanged, so you can see what the wording changes.

Worked example 4: proportion to a square root

Question. $y$ is directly proportional to $\sqrt{x}$ . When $x=4$ , $y=12$ . Work out $y$ when $x=16$ . Non-calculator.

Plan. Divide by $\sqrt4$ to find $k$ , then multiply by $\sqrt{16}$ .

\begin{aligned} y&=k\sqrt{x} &&\text{new model}\\ 12&=k\sqrt4=2k &&\text{evaluate the known root}\\ k&=12\div2=6 &&\text{find the constant}\\ y&=6\sqrt{16}=6\times4=24 &&\text{use the new input} \end{aligned}

Why does $y$ only double? Increasing $x$ by a factor of $4$ increases its square root by a factor of $2$ .

Answer. $y=24$ .

Check. $24\div\sqrt{16}=6=12\div\sqrt4$ , so the ratio to the root stays fixed.

You can often answer a multiplier question without finding $k$ . For $y=k\sqrt[3]{x}$ , multiplying $x$ by $8$ doubles $y$ , because $\sqrt[3]{8}=2$ . The unknown constant cancels from the comparison. When solving backwards in $y=kx^2$ , consider the domain: a positive physical length uses the positive square root, while unrestricted real $x$ may give two signs.

Watch out

Wrong: from $y\propto x^2$ , use $y=kx$ and square the final answer. Right: square $x$ inside the model before multiplying by $k$ . In general, $kx^2\ne(kx)^2$ .

This time the stated root is a cube root. The same find-the-constant method applies.

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R13 - Forming proportion equations (Higher)

Constructing a direct proportion equation (Higher)

Direct proportion to powers and roots (Higher)

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