The curve $x^2 + 3xy = x + 6$ and the line $2y - x = 2$ intersect at two points. Find the $x$-coordinates of the points of intersection.

1. Express $y$ in terms of $x$ using the linear equation

From $2y - x = 2$, we have:
$2y = x + 2$
$y = \frac{x + 2}{2}$

2. Substitute the expression for $y$ into the non-linear equation and simplify to a quadratic equation

Substitute $y = \frac{x + 2}{2}$ into $x^2 + 3xy = x + 6$:
$x^2 + 3x\left(\frac{x + 2}{2}\right) = x + 6$
Multiply the entire equation by 2 to remove the denominator:
$2x^2 + 3x(x + 2) = 2(x + 6)$
$2x^2 + 3x^2 + 6x = 2x + 12$
$5x^2 + 4x - 12 = 0$

3. Solve the quadratic equation to find the $x$-coordinates

Factorise the quadratic equation:
$(5x - 6)(x + 2) = 0$
$5x - 6 = 0 \implies x = \frac{6}{5} = 1.2$
$x + 2 = 0 \implies x = -2$