Algebraic proportion Worked Solutions — GCSE Maths

Turn the proportional statement into an equation

"Directly proportional to the square of $x$" becomes $y = k x^{2}$ once the constant $k$ is put in.

Substitute the pair of values given

Replace $x$ with $2$ and $y$ with $12$.

Solve for the constant

So the constant of proportionality is $k = 3$, and the model is $y = 3 x^{2}$.

Turn the proportional statement into an equation

"Directly proportional to the square of $x$" becomes $y = k x^{2}$ once the constant $k$ is put in.

Substitute the pair of values given

Replace $x$ with $5$ and $y$ with $50$.

Solve for the constant

So the constant of proportionality is $k = 2$, and the model is $y = 2 x^{2}$.

Turn the proportional statement into an equation

"Directly proportional to the cube of $x$" becomes $y = k x^{3}$ once the constant $k$ is put in.

Substitute the pair of values given

Replace $x$ with $2$ and $y$ with $24$.

Solve for the constant

So the constant of proportionality is $k = 3$, and the model is $y = 3 x^{3}$.

Turn the proportional statement into an equation

"Directly proportional to the square root of $x$" becomes $y = k \sqrt{x}$ once the constant $k$ is put in.

Substitute the pair of values given

Replace $x$ with $9$ and $y$ with $12$.

Solve for the constant

So the constant of proportionality is $k = 4$, and the model is $y = 4 \sqrt{x}$.

Turn the proportional statement into an equation

"Inversely proportional to the square of $x$" becomes $y = \frac{k}{x^{2}}$ once the constant $k$ is put in.

Substitute the pair of values given

Replace $x$ with $2$ and $y$ with $9$.

Solve for the constant

So the constant of proportionality is $k = 36$, and the model is $y = \frac{36}{x^{2}}$.