If the perimeter of the rectangle is 24cm, show that y = 9 − x, If the area of the rectangle is 27cm², show that x² − 6x = 0, Use your results in (i) and (ii) to find the width of the rectangle.

The perimeter $P$ of a rectangle with length $L$ and width $W$ is given by $P = 2(L+W)$.
Given $L = (x+3)$ cm and $W = y$ cm.
Given $P = 24$ cm.

Divide both sides by 2:

Subtract $x$ and 3 from both sides:

This shows the required relationship.

The area $A$ of a rectangle with length $L$ and width $W$ is given by $A = L \times W$.
Given $L = (x+3)$ cm and $W = y$ cm.
Given $A = 27$ cm$^2$.

Subtract 27 from both sides:

Multiply by -1:

This shows the required relationship.

Factor out $x$:

This gives two possible values for $x$:

If $x = 0$:

If $x = 6$:

Case 1: If $x = 0$
Length $L = (x+3) = (0+3) = 3$ cm.
Width $W = y = 9$ cm.
In this case, the "length" (3 cm) is shorter than the "width" (9 cm), which contradicts the usual convention that length is greater than or equal to width.

Case 2: If $x = 6$
Length $L = (x+3) = (6+3) = 9$ cm.
Width $W = y = 3$ cm.
In this case, the "length" (9 cm) is greater than the "width" (3 cm), which is consistent with the usual convention.

Based on the convention that the length of a rectangle is typically greater than or equal to its width, Case 2 is the valid solution.
Therefore, the width of the rectangle is 3 cm.

The width of the rectangle is $\boxed{\text{3 cm}}$.