Find the sum of the $x$-coordinates of the solutions to the system of equations $y=|x^2-6x+5|$ and $y=\frac{29}{4}-x$.
To find the solutions to this system, we need to find the values of x and y that make both equations true. Right. So let's look at the second equation. It tells us that y equals 29/4-x. So we can replace y with 29/4-x in the first equation. If we look at the left side of the equation, we see that $|x^2-6x+5|$ is always non-negative. That means that $\frac{29}{4}-x$ is also non-negative. We can find the values of x that make $\frac{29}{4}-x$ non-negative by solving $\frac{29}{4}-x\geq0$ So adding x to both sides we get $\frac{29}{4}\geq x$ That means that $x\leq\frac{29}{4}$. So we know that $x$ is $\frac{29}{4}$ or some number less than $\frac{29}{4}$. Now, let's figure out what values of $x$ make $x^2-6x+5=\frac{29}{4}-x$ true. If we add x to both sides, we get $x^2-5x+5=\frac{29}{4}$ We can multiply every term on both sides by 4 to get $4x^2-20x+20=29$ Subtracting 29 from both sides gives us $4x^2-20x-9=0$ This is a quadratic equation, so we can use the quadratic formula to solve for x: $x=\frac{-(-20)\pm\sqrt{(-20)^2-4(4)(-9)}}{2(4)}$ This simplifies to $x=\frac{5\pm\sqrt{34}}{2}$ Now, let's figure out what values of $x$ make $x^2-6x+5=-\frac{29}{4}+x$ true. We can start by subtracting x from both sides to get $x^2-7x+5=-\frac{29}{4}$. Multiplying every term on both sides by 4 gives us $4x^2-28x+20=-29$. Adding 29 to both sides we get $4x^2-28x+49=0$ This is a quadratic equation, so once again we can use the quadratic formula to solve for x: $x=\frac{-(-28)\pm\sqrt{(-28)^2-4(4)(49)}}{2(4)}$ It is $4*4*7^2-4(4)(49)=0$ So there is only one solution: $x=\frac{7}{2}$ So the possible values of $x$ are $\frac{7}{2}$, $\frac{5+\sqrt{34}}{2}$, and $\frac{5-\sqrt{34}}{2}$. $\frac{7}{2}$ is less than $\frac{29}{4}$, so it works. What about $\frac{5+\sqrt{34}}{2}$? That value is also less than $\frac{29}{4}$, so it is also a solution. Ok, and what about $\frac{5-\sqrt{34}}{2}$? That is less than $\frac{29}{4}$ as well, so all three values are valid. So we need to find the sum of $\frac{7}{2}$, $\frac{5+\sqrt{34}}{2}$, and $\frac{5-\sqrt{34}}{2}$. We can start by adding $\frac{7}{2}$ and $\frac{5+\sqrt{34}}{2}$ to get $\frac{7+5+\sqrt{34}}{2}$ This simplifies to $\frac{12+\sqrt{34}}{2}$ Now, let's add $\frac{12+\sqrt{34}}{2}$ and $\frac{5-\sqrt{34}}{2}$ to get $\frac{12+\sqrt{34}+5-\sqrt{34}}{2}$. This simplifies to $\frac{17}{2}$ So the sum of the x-coordinates is $\frac{17}{2}$.

