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If x, y, w, z are real numbers satisfying

 

w+x+y = -2

w+x+z = 4

w+y+z = 19

x+y+z = 12

 

What is wx + yz?

 Jun 15, 2021
 #1
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The solution is w = 1, x = -9, y = 6, z = 12, so wx + yz = (1)(-9) + (6)(12) = 63.

 Jun 15, 2021
 #2
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Sorry for this. The answer is actually $99$.Try adding all the equations together.

 Jun 15, 2021
 #3
avatar+26115 
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If x, y, w, z are real numbers satisfying
 
\(w+x+y = -2 \\ w+x+z = 4 \\ w+y+z = 19 \\ x+y+z = 12\)

 

What is \(wx + yz\)?

 

\(\begin{array}{|lrcll|} \hline (1)& w+x+y &=& -2 \\ (2)& w+x+z &=& 4 \\ (3)& w+y+z &=& 19 \\ (4)& x+y+z &=& 12 \\ \hline (1)+(2)+(3)+(4): & 3w+3x+3y+3z &=& -2+4+19+12 \\ & 3(w+x+y+z) &=& 33 \quad | \quad : 3 \\ & \mathbf{w+x+y+z} &=& \mathbf{11} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad x+y+z=12 \\ w+12 &=& 11 \\ w &=& 11-12 \\ \mathbf{w} &=& \mathbf{-1} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad w+y+z=19 \\ x+19 &=& 11 \\ x &=& 11-19 \\ \mathbf{x} &=& \mathbf{-8} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad w+x+z=4 \\ y+4 &=& 11 \\ y &=& 11-4 \\ \mathbf{y} &=& \mathbf{7} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{w+x+y+z} &=& \mathbf{11} \quad | \quad w+x+y=-2 \\ z-2 &=& 11 \\ z &=& 11+2 \\ \mathbf{z} &=& \mathbf{13} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline wx + yz &=& (-1)(-8)+7*13 \\ &=& 8+91 \\ \mathbf{wx + yz} &=& \mathbf{99} \\ \hline \end{array}\)

 

laugh

 Jun 15, 2021

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