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Let \(x\)\(y\), and \(z\) be real numbers such that

 

 \(\begin{align*} x+y-z &= -8, \\ x-y+z &= 18,\text{ and} \\ -x+y+z &= 30. \\ \end{align*}\)

 

Compute \(xyz\).

 Jul 7, 2022
edited by Guest  Jul 7, 2022
 #1
avatar+124676 
+2

Add the first two equations

 

2x =  10

x = 5

 

Add the second two

 

2z = 48

z = 24

 

Add the first and  third

 

2y = 22

y = 11

 

x y z =    

 

5 *  11  *  24   = 

 

1320

 

cool cool cool

 Jul 7, 2022
 #2
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Thank you very much, my good sir. Have a great day!

Guest Jul 7, 2022
 #3
avatar+124676 
0

You're welcome  !!!!

 

 

cool cool cool

CPhill  Jul 7, 2022

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