Find the exact roots of the equation $\left|\dfrac{4x-5}{x-2}\right| = 3x. On the same axes, sketch the curves with equations $y = \left|\frac{4x-5}{x-2}\right|$ and $y = 3x$. Hence solve exactly the inequality $\left|\frac{5-4x}{3x-6}\right| > x$.

1. Set up the two algebraic cases to remove the absolute value.

$\frac{4x-5}{x-2} = 3x \quad \text{or} \quad \frac{4x-5}{x-2} = -3x$

2. Solve the first case.

$4x - 5 = 3x(x - 2)$
$4x - 5 = 3x^2 - 6x$
$3x^2 - 10x + 5 = 0$
Using the quadratic formula:
$x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(5)}}{2(3)}$
$x = \frac{10 \pm \sqrt{40}}{6}$
$x = \frac{10 \pm 2\sqrt{10}}{6} = \frac{5 \pm \sqrt{10}}{3}$

3. Solve the second case.

$4x - 5 = -3x(x - 2)$
$4x - 5 = -3x^2 + 6x$
$3x^2 - 2x - 5 = 0$
$(3x - 5)(x + 1) = 0$
$x = \frac{5}{3} \quad \text{or} \quad x = -1$

4. Check for extraneous roots and conclude the final exact roots.

Since the absolute value must be non-negative, the right-hand side of the original equation requires $3x \ge 0 \implies x \ge 0$.
Therefore, $x = -1$ is an extraneous root and is rejected.
The remaining roots are positive and valid.
$x = \frac{5}{3}, \frac{5 - \sqrt{10}}{3}, \frac{5 + \sqrt{10}}{3}$

1. Compute the exact points of intersection between the two graphs (typically done in part (i)).

Solve $\left|\frac{4x-5}{x-2}\right| = 3x$.
Since the left-hand side is non-negative, $x \ge 0$.

Case 1: $\frac{4x-5}{x-2} = 3x$
$4x - 5 = 3x^2 - 6x \implies 3x^2 - 10x + 5 = 0$
$x = \frac{10 \pm \sqrt{100 - 60}}{6} = \frac{5 \pm \sqrt{10}}{3}$

Case 2: $\frac{4x-5}{x-2} = -3x$
$4x - 5 = -3x^2 + 6x \implies 3x^2 - 2x - 5 = 0$
$(3x - 5)(x + 1) = 0 \implies x = \frac{5}{3} \text{ or } x = -1$
Reject $x = -1$ since we require $x \ge 0$. The exact intersections occur at $x = \frac{5-\sqrt{10}}{3}, \frac{5}{3}, \frac{5+\sqrt{10}}{3}$.

2. Identify key features and sketch the graphs of $y = \left|\frac{4x-5}{x-2}\right|$ and $y = 3x$.

For $y = \left|\frac{4x-5}{x-2}\right|$, rewrite the rational function as $y = \left|4 + \frac{3}{x-2}\right|$.

• Vertical asymptote: $x = 2$
• Horizontal asymptote: $y = 4$
• $x$-intercept: $x = \frac{5}{4} = 1.25$
• $y$-intercept: $y = \frac{5}{2} = 2.5$

The absolute value reflects the portion below the $x$-axis (between $x = 1.25$ and $x = 2$) upwards, meaning as $x \to 2^-$, the curve goes to $+\infty$. As $x \to 2^+$, the curve also comes from $+\infty$ and approaches $y = 4$.

The line $y = 3x$ passes through $(0,0)$ and intersects the curve at the $3$ points calculated in Step 1.
(Student draws axes, labels intercepts and asymptotes, and accurately sketches the line and the reflected hyperbola.)

3. Manipulate the given inequality to relate it directly to the sketched graphs.

Given the inequality:
$\left|\frac{5-4x}{3x-6}\right| > x$

Factor out constants to match the original function:
$\left|\frac{-(4x-5)}{3(x-2)}\right| > x$
$\frac{|-(4x-5)|}{|3(x-2)|} > x$
$\frac{1}{3}\left|\frac{4x-5}{x-2}\right| > x$
$\left|\frac{4x-5}{x-2}\right| > 3x$

4. Solve the inequality by identifying regions where the curve is above the line.

We need the intervals of $x$ where the curve $y = \left|\frac{4x-5}{x-2}\right|$ lies above the line $y = 3x$.
Observing the sketch and intersection points from left to right:

• For $x < \frac{5-\sqrt{10}}{3}$, the curve is above the line.
• For $\frac{5-\sqrt{10}}{3} < x < \frac{5}{3}$, the line is above the curve.
• For $\frac{5}{3} < x < \frac{5+\sqrt{10}}{3}$, the curve goes to $+\infty$ near $x = 2$, placing it above the line. However, the curve is undefined exactly at the asymptote $x = 2$.
• For $x > \frac{5+\sqrt{10}}{3}$, the line overtakes the curve and remains above it.

Hence, the exact solution is $x < \frac{5-\sqrt{10}}{3} \text{ or } \frac{5}{3} < x < 2 \text{ or } 2 < x < \frac{5+\sqrt{10}}{3}$.