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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.

 Jul 6, 2023
 #1
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0

Adding the last three equations, we get [(p + q + r) + (q + r)x + (p + r)y + (p + q)z = x + y + z.]Substituting p+q+r=−3, we get −3+(q+r)x+(p+r)y+(p+q)z=x+y+z.

Manipulating the first equation and substituting, we get \begin{align*} -3 + (-3 - p)x + (-3 - q)y + (-3 - r)z &= x + y + z \ \Rightarrow \quad -3 - px - qy - rz &= x + 3x + y + 3y + z + 3z \ \Rightarrow \quad -3 - (px + qy + rz) &= 4 (x + y + z) \ \Rightarrow \quad x + y + z &= \frac{-2}{4} = -\frac{1}{2}. \end{align*}

Therefore, x+y+z=−1/2.​​

 Jul 6, 2023
 #2
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0

Nice try, but that's not correct...

 Jul 6, 2023
 #3
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It's actually -1

 Jul 6, 2023
 #4
avatar+397 
+1

Add the top four equations and you get

\(p(1+x+y+z)+q(1+x+y+z)+r(1+x+y+z)=1+x+y+z\)

so

\(p+q+r=1,\)

implying that the equations are inconsistent.

 Jul 7, 2023

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