Find the sum of all the values of $x$ that satisfy the system of equations $y=|x^2-6x+5|$ and $y=\frac{29}{4}-2x$.
First we can examine the graph of $y=|x^2-6x+5|$ to get an idea of how it behaves.
[asy]
Label f;
f.p=fontsize(4);
xaxis(-5,7,Ticks(f, 1.0));
yaxis(-3,5,Ticks(f, 1.0));
real f(real x)
{
return abs(x^2-6*x+5);
}
draw(graph(f,-3,5),blue);
[/asy]
Note that $y=|x^2-6x+5|$ is just the graph of $y=x^2-6x+5$ reflected across the $x$-axis for $x\in(1,5)$, so to find the solutions to the system, we must find the $x$ such that $$x^2-6x+5=\frac{29}{4}-2x\qquad\Rightarrow\qquad x^2-4x-\frac{11}{4}=0.$$We can factor the left hand side and then apply the quadratic formula to get that $$x=\frac{4\pm\sqrt{16+11}}{2}=\frac{4\pm\sqrt{27}}{2}.$$Hence, the two solutions are $x=2+\sqrt{27}/2$ and $x=2-\sqrt{27}/2$. The sum of the two solutions is $\boxed{4}$.
To find the sum of all the values of $x$ that satisfy the system of equations, we need to find the values of $x$ that make both equations true simultaneously.
Let's solve the equations step by step:
Step 1: Set the equations equal to each other:
\[|x^2-6x+5| = \frac{29}{4}-2x\]
Step 2: Rewrite the equation using absolute value definition:
\[(x^2-6x+5) = \frac{29}{4}-2x \quad \text{or} \quad (x^2-6x+5) = -\left(\frac{29}{4}-2x\right)\]
Step 3: Solve both equations separately:
For the first case:
\[x^2-6x+5 = \frac{29}{4}-2x\]
Step 4: Move all terms to one side to form a quadratic equation:
\[x^2 - 6x + 5 - \frac{29}{4} + 2x = 0\]
Step 5: Combine like terms:
\[x^2 - 4x + \frac{11}{4} = 0\]
Step 6: Multiply the equation by $4$ to eliminate fractions:
\[4x^2 - 16x + 11 = 0\]
Step 7: Solve the quadratic equation using factoring or the quadratic formula. In this case, it cannot be factored easily, so we will use the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Applying the quadratic formula, we get:
\[x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(4)(11)}}{2(4)}\]
Simplifying the expression gives us:
\[x = \frac{16 \pm \sqrt{256 - 176}}{8}\]
\[x = \frac{16 \pm \sqrt{80}}{8}\]
\[x = \frac{16 \pm \sqrt{16 \cdot 5}}{8}\]
\[x = \frac{16 \pm 4\sqrt{5}}{8}\]
\[x = \frac{4}{2} \pm \frac{\sqrt{5}}{2}\]
Therefore, in the first case, we have two solutions for $x$:
\[x_1 = 2 + \frac{\sqrt{5}}{2} = 2 + 0.5\sqrt{5} \approx 4.14\]
\[x_2 = 2 - \frac{\sqrt{5}}{2} = 2 - 0.5\sqrt{5} \approx -0.14\]
For the second case, we have:
\[x^2 - 6x + 5 = -\left(\frac{29}{4}-2x\right)\]
Following the same steps as before, we get the equation:
\[4x^2 - 20x + \frac{59}{4} = 0\]
Solving this equation using the quadratic formula gives us the other two solutions for $x$:
\[x_3 = \frac{5 + \sqrt{5}}{2} \approx 3.62\]
\[x_4 = \frac{5 - \sqrt{5}}{2} \approx 1.38\]
Now, that we have all the solutions for $x$, we can find their sum:
\[x_1 + x_2 + x_3 + x_4 = (2 + 0.5\sqrt{5}) + (2 - 0.5\sqrt{5}) + (\frac{5 + \sqrt{5}}{2}) + (\frac{5 - \sqrt{5}}{2})\]
\[= 4 + \frac{10}{2}\]
\[= 4 + 5\]
\[= 9\]
Therefore, the sum of all the values of $x$ that satisfy the given system of equations is $9$.