Find the sum of the $x$-coordinates of the solutions to the system of equations $y=|x^2-6x+5|$ and $y=\frac{29}{4}-x$.
The quadratic $x^2-6x+5$ factors as $(x-5)(x-1)$, so it crosses the $x$-axis at $1$ and $5$. Since the leading coefficient is positive, it opens upwards, and thus the value of the quadratic is negative for $x$ between $1$ and $5$. Thus if $x\le 1$ or $x\ge 5$, we have $|x^2-6x+5|=x^2-6x+5$. We can solve the system in this range by setting the $y$-values equal, so

\begin{align*}
x^2-6x+5&=\frac{29}{4}-x\\
x^2-5x+\frac{20}{4}-\frac{29}{4}&=0\\
x^2-5x-\frac{9}{4}&=0.
\end{align*}Thus by the quadratic formula, $$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(\frac{-9}{4})(1)}}{2(1)}=\frac{5\pm\sqrt{25+9}}{2}=\frac{5\pm\sqrt{34}}{2}.$$A quick check shows that both solutions have either $x<1$ or $x>5$, so they are both valid in this system. We do not need to find the corresponding $y$-values since the problem asks only for the sum of the $x$-coordinates.

If $1\le x\le 5$, we know $|x^2-6x+5|=-x^2+6x-5$. Solving the system as before, we have

\begin{align*}
\frac{29}{4}-x&=-x^2+6x-5\\
x^2-7x+\frac{29}{4}+\frac{20}{4}&=0\\
x^2-7x+\frac{49}{4}&=0\\
(x-\frac{7}{2})^2&=0\\
x&=\frac{7}{2}.
\end{align*}Checking, this value is indeed between $1$ and $5$, so it is allowable. Thus the possible $x$-values are $\frac{5+\sqrt{34}}{2}$, $\frac{5-\sqrt{34}}{2}$, and $\frac{7}{2}$. Their sum is $$\frac{5+\sqrt{34}}{2}+\frac{5-\sqrt{34}}{2}+\frac{7}{2}=\frac{5+5+7}{2}=\boxed{\frac{17}{2}}.$$
\frac{17}{2}
Level 5
Algebra
Problem: Find the sum of the $x$-coordinates of the solutions to the system of equations $y=|x^2-6x+5|$ and $y=\frac{29}{4}-x$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the sum of the $x$-coordinates of the solutions, first, we need to analyze the equations given. The first equation, $y=|x^2-6x+5|$, can be rewritten by factoring the quadratic expression inside the absolute value. The expression $x^2-6x+5$ factors to $(x-1)(x-5)$. The roots of this quadratic are $x=1$ and $x=5$. 

Next, we determine the behavior of the quadratic. The parabola opens upwards, so it is below the x-axis between its roots, specifically for $1 < x < 5$. Therefore, we can express the absolute value function as:
- $y = x^2 - 6x + 5$ for $x \leq 1$ and $x \geq 5$.
- $y = -(x^2 - 6x + 5) = -x^2 + 6x - 5$ for $1 < x < 5$.

The second equation, $y=\frac{29}{4}-x$, is a linear equation with a negative slope. To find the intersection points, we set the two equations equal to each other.

1. For $x \leq 1$ and $x \geq 5$, we solve:
   $x^2 - 6x + 5 = \frac{29}{4} - x$.
   Rearranging gives:
   $x^2 - 5x + 5 - \frac{29}{4} = 0$.
   Multiplying through by 4 to eliminate the fraction results in:
   $4x^2 - 20x + 20 - 29 = 0$,
   which simplifies to:
   $4x^2 - 20x - 9 = 0$.
   Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find the roots.

2. For $1 < x < 5$, we solve:
   $-x^2 + 6x - 5 = \frac{29}{4} - x$.
   Rearranging gives:
   $-x^2 + 7x - 5 - \frac{29}{4} = 0$.
   Multiplying through by 4 results in:
   $-4x^2 + 28x - 20 - 29 = 0$,
   which simplifies to:
   $-4x^2 + 28x - 49 = 0$.
   Again, we apply the quadratic formula to find the roots.

After calculating the roots from both cases, we sum the $x$-coordinates of all solutions. The final result of this summation yields $\frac{17}{2}$.<|END_OF_LATENT|> The final answer is $\frac{17}{2}$.