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# Help

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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.

Oct 10, 2019

#1
+2

Here is a hint:

See the equations that have isolated x, y, z?

Now, notice the other equations, how can you susbtitute in and evualate?

Its pretty simple, you just have to see it!

Oct 10, 2019
#2
+2

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

CalculatorUser  Oct 10, 2019
#4
+2

Thanks for that hint, CU......that's pretty crafty   !!!   CPhill  Oct 10, 2019
#3
0

I am lowkey confused cause these variables are hard to combine. Coul you help me some more?

Oct 10, 2019
#5
+2

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

I think so

Oct 10, 2019
edited by CalculatorUser  Oct 10, 2019
edited by 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   Oct 10, 2019
#7
+2

Yes, this is correct!

CalculatorUser  Oct 10, 2019
#8
+1

Well, CU......your hint was a GREAT one   !!!

I wouldn't have seen how to proceed without it  !!!   CPhill  Oct 10, 2019
edited by CPhill  Oct 10, 2019
#9
+1

Tahnks!

CalculatorUser  Oct 10, 2019
#10
+1

Thanks Boys!!

Oct 10, 2019