Must be Algebra

$\begin{cases} x+7y+3v+5u=16&\implies\boxed{1}\\ 8x+4y+6v+2u=-16&\implies\boxed{2}\\ 2x+6y+4v+8u=16&\implies\boxed{3}\\ 5x+3y+7v+1u=-16&\implies\boxed{4}\\ \end{cases}$

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

$6x+10y+10v+6u=0\implies\boxed{5}$

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

$10x+10y+10v+10u=0\implies\boxed{6}$

$\boxed{6}-\boxed{5}:$

$4x+4u=0\\ x+u=0\implies\boxed{7}\\ x=-u \implies\boxed{8}$

Substitute $\boxed{7}$ into $\boxed{5}:$

$6(x+u)+10(y+v)=0\\ 10(y+v)=0\\ y+v=0\\ v=-y \implies\boxed{9}$

Substitute $\boxed{8}$ and $\boxed{9}$ into $\boxed{1}:$

$-u+7y+3(-y)+5u=16\\ 4u+4y=16\\ u+y=4\implies\boxed{10}$

Substitute $\boxed{8}$ and $\boxed{9}$ into $\boxed{3}:$

$2(-u) + 6y+4(-y)+8u = 16\\ 6u+2y=16\\ 3u+y=8 \implies \boxed{11}$

$\boxed{11} - \boxed{10}:$

$2u=4 \implies u=2\\ 2+y=4 \implies y=2\\ v=-y=-2\\ x=-u=-2$

Therefore, $xyuv = -2(2)(2)(-2) = \color{#D61F06}{\boxed{\color{#333333}{16}}}$

Easiest method!

Put the left half of the equation into a matrix (not the variables, the coefficients) (4 by 4), take the inverse, then put the right half into a matrix (4 by 1) and multiply the two together. You will get a matrix (4 by 1) where the top number is x, the next is y... (in the future order try appear in the equations). You should have 2, -2, 2, -2. Multiplied together, it is 16.

This sounds complicated, but once you understand it and you use a calculator, it is quite simple. In fact, I did this on my phone! (So sorry if there are typos in this solution!)

$\begin{cases} x + 7y + 3v + 5u = 16 & ...(1) \\ 8x + 4y + 6v + 2u = -16 & ...(2) \\ 2x + 6y + 4v + 8u = 16 & ...(3) \\ 5x + 3y + 7v + u = -16 & ...(4) \end{cases}$

$\begin{array} {lrl} (1)+(2): & 9x + 11y + 9v + 7u & = 0 & ...(5) \\ (3)+(4): & 7x + 9y + 11v + 9u & = 0 & ...(6) \\ (3)-(1): & x - y + v + 3u & = 0 & ...(7) \\ (2)-(4): & 3x + y - v + u & = 0 & ...(8) \end{array}$

$\begin{aligned} (7)+(8): \quad 4x + 4u & = 0 \\ \implies x & = -u \end{aligned}$

$\begin{aligned} (5)+(6): \quad 16x + 20y + 20v + 16u & = 0 \\ 20y + 20v & = 0 \\ \implies y & = -v \end{aligned}$

$\begin{aligned} (8): \quad 3x + y - v + u & = 0 \\ 3x + y + y -x & = 0 \\ 2x+2y & = 0 \\ \implies x & = -y = v = -u \end{aligned}$

$\begin{aligned} (1): \quad x + 7y + 3v + 5u & = 16 \\ x - 7x + 3x - 5x & = 16 \\ -8x & = 16 \\ \implies x & = -2 \end{aligned}$

$\implies x = -2, \ y = 2, \ v = -2, \ u = 2, \ \implies xyuv = \boxed{16}$

$\begin{cases} x + 7y + 3v + 5u = 16 \\ 8x + 4y + 6v + 2u = -16 \\ 2x + 6y + 4v + 8u = 16 \\ 5x + 3y + 7v + u = -16 \end{cases}$

Real numbers $x$ , $y$ , $u$ , and $v$ satisfy the system of equations above. Find $xyuv$ .

Eq 1 + Eq 4

= 6 $x$ + 10 $y$ + 10 $v$ + 6 $u$ = 0

Eq 2 + Eq 3

= 10 $x$ + 10 $y$ + 10 $v$ + 10 $u$ = 0

From this we obtain

6 ( $x + u$ ) + 10 ( $y + v$ ) = 0

10 ( $x + u$ ) + 10 ( $y + v$ ) = 0

which yields $x + u$ = 0 and $y + v$ = 0.

Substituting - $x$ for $u$ , and $-y$ for $v$ in the original equation, we find

$- 4x + 4y$ = 16

$6x - 2y$ = -16

$2x + 2y$ = 0

Plugging this in with earlier assumptions we find

$x$ = -2, $y$ = 2, $u$ = 2, $v$ = -2

$xyuv$ = 16

You can also solve this using a matrix (4X4 matrices are tedious to invert by hand but a computer makes short work of it).

you can solve it using matrix multiplication, $A*B = C$ then $B = A^-1*C$ were A is the coefficients matrix, B is the variables matrix, and C is the results matrix. see https: //www.mathsisfun.com/algebra/systems-linear-equations-matrices.html for further explanation.