Find the value of $r$ such that \[\frac{r^2 - 5r + 4}{r^2-8r+7} = \frac{r^2 - 2r -15}{r^2 + r - 12}.\]
We set the fractions equal to each other and cross-multiply: \[\begin{aligned} (r^2 - 5r + 4)(r^2 + r - 12) &= (r^2 - 2r - 15)(r^2-8r+7)\\ r^4 + 2r^3 - 8r^2 - 10r^3 - 20r^2 + 80r - 60r - 48 &= r^4 - 8r^3 + 7r^2 -2r^3 +16r^2 -15r^2 - 120r + 112r - 105\\ r^4 - 8r^3 - 3r^2 - 104r + 57 &= 0. \end{aligned}\]If the quadratic $r^2 - 8r - 3$ has any rational roots, they must be divisors of 3. We check 1, -1, 3, and -3, and quickly find that only $\boxed{r = -1}$ and $r = 3$ are solutions.
To find the value of $r$, we can start by multiplying both sides of the equation by the denominators of the fractions to clear the fractions.
For the left-hand side, we have \[\frac{r^2 - 5r + 4}{r^2-8r+7} = \frac{(r-1)(r-4)}{(r-1)(r-7)}.\]
And for the right-hand side, we have \[\frac{r^2 - 2r -15}{r^2 + r - 12} = \frac{(r-5)(r+3)}{(r+4)(r-3)}.\]
Hence, our equation becomes \[\frac{(r-1)(r-4)}{(r-1)(r-7)} = \frac{(r-5)(r+3)}{(r+4)(r-3)}.\]
Now we can cross-multiply to obtain \[(r-1)(r-4)(r+4)(r-3) = (r-1)(r-7)(r-5)(r+3).\]
Next, we can simplify the equation by canceling common factors on both sides.
We have \[(r-4)(r+4)(r-3) = (r-7)(r-5)(r+3).\]
Expanding both sides gives \[(r^2-16)(r-3) = (r^2-2r-35)(r+3).\]
Expanding further, we have \[r^3 - 19r^2 + 48r - 48 = r^3 + r^2 - 37r - 105.\]
Simplifying this equation further, we obtain \[-19r^2 + 48r - 48 = r^2 - 37r - 105.\]
Combining like terms gives \[0 = 20r^2 - 85r - 57.\]
Now, we can solve for $r$ by factoring or using the quadratic formula.