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If $w$, $x$, $y$, and $z$ are real numbers satisfying:\begin{align*} w+x+y &= -2, \\ w+x+z &= 4, \\ w+y+z &= 19, \text{ and} \\ x+y+z &= 12, \end{align*}what is $wx + yz$?

 Apr 12, 2021
 #1
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Add  the  first  three equations and we  get  that

 

2x + 2y + 2z  + 3w  =  21

 

2 ( x +y + z)  +  3w  =  21                 sub in  the  last equation for  x + y + z

 

2 (12)  + 3w  =  21

 

3w  =  21  - 24

 

3w   = -3

 

w =  -1

 

Subtract  the second equation from the first

y - z =  -6     ( 4)

 

And using the  third  equation

y + z   =  19 - w

y + z  =  19 - - 1

y + z  =  20   (5)

 

Add (4)  and (5)

2y  =  14

y  =  7

 

So 

 y +  z  =  20

7  + z  = 20

z  = 13

 

And

x + y + z  = 12

x + 7 + 13  = 12

x + 20  = 12

x  =  -8

 

wx  + yz    =    (-1) (-8)  +  (7) ( 13)   =   8  +  91   =    99

 

 

cool cool cool

 Apr 12, 2021
edited by CPhill  Apr 12, 2021

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