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.

Logic Oct 10, 2019

#1**+2 **

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!

CalculatorUser Oct 10, 2019

#2**+2 **

You can ask for the explanation and solution if you are having trouble.

CalculatorUser
Oct 10, 2019

#3**0 **

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?

Logic Oct 10, 2019

#5**+2 **

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

CalculatorUser Oct 10, 2019

#6**+2 **

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

CPhill Oct 10, 2019