Birthday Cake

Let the number of students be $x$ and the cost of the cake be $y$ .

$2x = y + 4 - ①$

$3x = 2y - ②$

From ①,

$y = 2x - 4$

Substitute $y$ into ②,

$3x = 2 (2x - 4)$

We can work out that $x = 8$ .

Substitute $x$ back into any equation, let's say ①,

$2 (8) = y + 4$

Therefore $y = 12$ .

The cost of the cake = $\$12$

They said that If each of them paid $\$3$ then there would be enough for $2$ cakes. So, each of them needs to pay $\$3 \div 2 = \$1.5$ for it to be exactly enough to buy a cake.

So the next steps would be: $\dfrac{\$4}{\$2-\$1.5} \implies 8 pupils$

Since our first statement is that each needs to pay $\$1.5$ to get exactly 1 cake.

$8\times \$1.5 \longrightarrow \$12$

Let $x$ be the number of students and $c$ be the cost of one cake in $. Then we have the equations,

$2x=c+4$ $\color{#D61F06}(1)$

$3x=2c$ $\color{#D61F06}(2)$

Solve for $x$ in terms of $c$ in $\color{#D61F06}(2)$ then substitute in $\color{#D61F06}(1)$ . We have

$2\left(\dfrac{2}{3}c\right)=c+4$ $\implies$ $\dfrac{4}{3}c-c=4$ $\implies$ $c=\boxed{12}$