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 $wx + yz$?
+128474 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
