Given

\begin{align*}

px+qy+rz&=1,\\

p+qx+ry&=z,\\

pz+q+rx&=y,\\

py+qz+r&=x,\\

p+q+r&=-3,

\end{align*}

find $x+y+z$.

wizzymath Mar 10, 2021

#1**0 **

Yikes! This looks ugly, but I think I cracked this one after some deep thinking. Let me show you my working. This one is quite intimidating, but some key observations and some determination should be enough to figure out the answer to this question successfully.

I started by adding x + y + z and seeing if I could get anywhere with adding the three equations together. I cleverly grouped terms together to make some patterns clearer to me as I solved.

\(x + y + z =(py + qz + r) + (pz + q + rx) + (p + qx + ry)\\ x + y + z = (p + q + r) + (py + pz) + (qx + qz) + (rx + ry)\)

I stopped here to reflect on my strategy thus far. While it certainly looks like I have made this much worse, I noticed that p+q+r = -3, so I could at least do that substitution. I then also noticed that px + qy + rz = 1 would maybe be useful because it would allow me to factor out another factor of p+q+r again. Let's show this in action. The answer just falls out on its own without much effort after that.

\(x + y + z = -3 + (py + pz) + (qx + qz) + (rx + ry)\\ x + y + z + 1 = -3 + (py + pz) + (qx + qz) + (rx + ry) + px + qy + rz\\ x + y + z = -4 + (px + py + pz) + (qx + qy + qz) + (rx + ry + rz)\\ x + y + z = -4 + p(x + y + z) + q(x + y + z) + r(x + y + z)\\ x + y + z = -4 + (p + q + r)(x + y + z)\\ x + y + z = -4 - 3(x + y + z)\\ 4(x + y + z)= -4\\ x + y + z = -1\)

Guest Mar 10, 2021