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.
Here is a hint:
See the equations that have isolated x, y, z?
Add them up.
Now, notice the other equations, how can you susbtitute in and evualate?
Its pretty simple, you just have to see it!
You can ask for the explanation and solution if you are having trouble.
so P+qx+ry+pz+q+rx+py+qz+r = the answer
I am lowkey confused cause these variables are hard to combine. Coul you help me some more?
your close!
rearrange and subtitute in -3
(qx + ry + pz) + (rx + py + qz) - 3 = x + y + z
Oops I realized the next step is wrong! The problem may be harder than I thought!
-3 ( x + y + z) + (-3) (x + y + z) - 3 = x + y + z (undistribute p + q + r because they = -3)
-3x - 3y - 3z -3x - 3y - 3z - 3 = x + y + z
-6x - 6y - 6z - 3 = x + y + z
7x + 7y + 7z = 3
answer is 3/7
I think so
Using CU's hint :
(p + q+ r) + (qx + ry) + (pz + rx) + ( py + qz) = x + y + z
-3 + x (q + r) + y (p + r) + z(p + q) = x + y + z
-3 + x(-3 - p) + y (-3 - q) + z (-3 - r) = x + y + z
-3 - 3x - px - 3y - qy - 3z - rz = x + y + z
-3 - 3 ( x + y + z) - ( px + qy + rz) = x + y + z
-3 - 3(x + y + z) - (1) = x + y + z
-3 - 1 = 4 (x + y + z)
-4 = 4 (x + y + z)
-1 = x + y + z