Here are two right-angled triangles - Edexcel - GCSE Maths - Question 21 - 2018 - Paper 3

Since tan e = tan f, we have: $\frac{x}{4x - 1} = \frac{6x + 5}{12x + 31}$

Cross multiplying gives: $(x)(12x + 31) = (6x + 5)(4x - 1)$

Distributing each side: $12x^2 + 31x = 24x^2 - 6x + 20x - 5$

This simplifies to: $12x^2 + 31x = 24x^2 + 14x - 5$

Moving all terms to one side results in: $12x^2 - 24x^2 + 31x - 14x + 5 = 0$

Which simplifies to: $-12x^2 + 17x + 5 = 0$

Or, multiplying through by -1: $12x^2 - 17x - 5 = 0$

Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 12$, $b = -17$, and $c = -5$:

First find the discriminant: $b^2 - 4ac = (-17)^2 - 4(12)(-5) = 289 + 240 = 529$

Then substituting into the formula: $x = \frac{17 \pm \sqrt{529}}{24} = \frac{17 \pm 23}{24}$

Calculating the two possible values:

1. $x = \frac{40}{24} = \frac{5}{3}$
2. $x = \frac{-6}{24} = -\frac{1}{4}$

Since $x$ must be positive, we select: $x = \frac{5}{3}$

$x = \frac{5}{3}$, or approximately $1.67$.