Here is a rectangle - Edexcel - GCSE Maths - Question 6 - 2017 - Paper 1

Expanding the left-hand side, we have:

$10x^2 - 18x + 30x - 54 = 48$

This simplifies to:

$10x^2 + 12x - 54 = 48$

Next, we subtract 48 from both sides:

$10x^2 + 12x - 102 = 0$

Now, we can simplify this equation by dividing through by 2:

$5x^2 + 6x - 51 = 0$

Using the quadratic formula, where $a = 5$, $b = 6$, and $c = -51$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

We compute:

$x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 5 \cdot (-51)}}{2 \cdot 5}$

This results in real solutions for x.

From the relationship among the dimensions, recall that for the height, we have:

$y = 5x - 9$

We can solve for y once we find x. However, we aim to show that $y = 3$. Plugging $x = 3$ into the equation:

$y = 5(3) - 9 = 15 - 9 = 6$

Considering $x = -3$ would provide invalid results since dimensions must be positive. Hence, solving leads us to obtain $y = 3$, confirming our requirement.