Every Equation Is Missing Something

Relevant wiki: Systems of Linear Equations - Substitution

$\begin{cases}x+y+u=4 &\implies\boxed{1}\\ y+u+v = -5&\implies\boxed{2}\\ u+v+x=0&\implies\boxed{3}\\ v+x+y=-8 &\implies\boxed{4}\end{cases}$

$\boxed{1}+\boxed{2}+\boxed{3}+\boxed{4}$ :

$3x+3y+3u+3v = 4+(-5)+0+(-8)\\ 3(x+y+u+v) = -9\\ x+y+u+v = -3 \implies\boxed{5}$

Substitute $\boxed{1}$ into $\boxed{5}$ :

$4+v = -3 \implies v=-7$

Substitute $\boxed{2}$ into $\boxed{5}$ :

$-5+x= -3 \implies x=2$

Substitute $\boxed{3}$ into $\boxed{5}$ :

$0+y= -3 \implies y=-3$

Substitute $\boxed{4}$ into $\boxed{5}$ :

$-8+u= -3 \implies u=5$

Therefore, $xyuv = (2)(-3)(5)(-7) = \boxed{210}$

$\begin{cases} u + 0 + x + y = 4 & ... (1) \\ u + v + 0 + y = -5 & ... (2) \\ u + v + x + 0 = 0 & ... (3) \\ 0 + v + x + y = -8 & ... (4) \end{cases}$

$\begin{aligned} (1)+(2)+(3)+(4): \quad 3(u+v+x+y) & = - 9 \\ \implies u+v+x+y & = - 3 \end{aligned}$

Now, consider:

$\begin{aligned} (1): \quad u + 0 + x + y & = 4 \\ u + v + x + y - v & = 4 \\ -3 - v & = 4 \\ \implies v & = - 7 \end{aligned}$

Similarly,

$\begin{aligned} (2): \quad u + v + 0 + y & = -5 \\ \implies x & = 2 \end{aligned}$

$\begin{aligned} (3): \quad u + v + x + 0 & = 0 \\ \implies y & = -3 \end{aligned}$

$\begin{aligned} (4): \quad 0 + v + x + y & = -8 \\ \implies u & = 5 \end{aligned}$

$\implies xyuv = (2)(-3)(5)(-7) = \boxed{210}$