The curve $x^2 + xy - 6y^2 + 9 = 0$ and line $y = x + 1$ intersect at the points $R$ and $S$. Find the $x$-coordinate of the midpoint of $RS$.

1. Substitute the linear equation into the curve equation

Substitute $y = x + 1$ into $x^2 + xy - 6y^2 + 9 = 0$:
$x^2 + x(x + 1) - 6(x + 1)^2 + 9 = 0$

2. Expand and simplify to form a quadratic equation in $x$

$x^2 + x^2 + x - 6(x^2 + 2x + 1) + 9 = 0$
$2x^2 + x - 6x^2 - 12x - 6 + 9 = 0$
$-4x^2 - 11x + 3 = 0$
$4x^2 + 11x - 3 = 0$

3. Solve for $x$ and find the $x$-coordinate of the midpoint

$(4x - 1)(x + 3) = 0$
$x = \frac{1}{4} \text{ or } x = -3$
The $x$-coordinate of the midpoint is the average of the $x$-coordinates of $R$ and $S$:
$x_m = \frac{\frac{1}{4} + (-3)}{2} = \frac{-\frac{11}{4}}{2} = -\frac{11}{8}$