The line $2y = 3x - 7$ meets the curve $y = x^2 + 3x - 8$ at two points $A$ and $B$.
Find the distance between $A$ and $B$.

1. Substitute the linear equation into the quadratic equation

From $2y = 3x - 7$, we have $y = \frac{3x - 7}{2}$.
Substitute this into $y = x^2 + 3x - 8$:
$\frac{3x - 7}{2} = x^2 + 3x - 8$

2. Simplify to form a quadratic equation in $x$

$3x - 7 = 2(x^2 + 3x - 8)$
$3x - 7 = 2x^2 + 6x - 16$
$2x^2 + 3x - 9 = 0$

3. Solve the quadratic equation for $x$

$(2x - 3)(x + 3) = 0$
$x = \frac{3}{2} \text{ or } x = -3$

4. Find the corresponding $y$-coordinates

When $x = \frac{3}{2}$, $y = \frac{3(\frac{3}{2}) - 7}{2} = -\frac{5}{4}$
When $x = -3$, $y = \frac{3(-3) - 7}{2} = -8$
The points of intersection are $A\left(\frac{3}{2}, -\frac{5}{4}\right)$ and $B(-3, -8)$.

5. Calculate the distance between the two points

$AB = \sqrt{\left(\frac{3}{2} - (-3)\right)^2 + \left(-\frac{5}{4} - (-8)\right)^2}$
$AB = \sqrt{\left(\frac{9}{2}\right)^2 + \left(\frac{27}{4}\right)^2}$
$AB = \sqrt{\frac{81}{4} + \frac{729}{16}} = \sqrt{\frac{1053}{16}}$
$AB \approx 8.1125 \approx 8.11 \text{ units (to 3 s.f.)}$