Okay, I will solve the systems of linear equations by elimination and match the solutions to the corresponding words. Then, I will answer the synthesis question.

Task 1: THE

The system of equations is:
$7x - 5y = 26$
$2x + y = 5$

Multiply the second equation by 5 to eliminate $y$:
$10x + 5y = 25$

Add the modified second equation to the first equation:
$(7x - 5y) + (10x + 5y) = 26 + 25$
$17x = 51$
$x = \frac{51}{17} = 3$

Substitute $x = 3$ into the second equation:
$2(3) + y = 5$
$6 + y = 5$
$y = 5 - 6 = -1$

The solution is $(3, -1)$. This solution is not in the Answer Box. Let's check the calculations.

$7(3) - 5(-1) = 21 + 5 = 26$
$2(3) + (-1) = 6 - 1 = 5$

The solution $(3, -1)$ is correct. There might be a typo in the Answer Box.

Task 2: STUFFED

The system of equations is:
$-4x + 9y = 9$
$x - 3y = -6$

Multiply the second equation by 4 to eliminate $x$:
$4x - 12y = -24$

Add the modified second equation to the first equation:
$(-4x + 9y) + (4x - 12y) = 9 + (-24)$
$-3y = -15$
$y = \frac{-15}{-3} = 5$

Substitute $y = 5$ into the second equation:
$x - 3(5) = -6$
$x - 15 = -6$
$x = -6 + 15 = 9$

The solution is $(9, 5)$.

Task 3: ARE

The system of equations is:
$3x + 2y = 12$
$x - 2y = 4$

Add the two equations to eliminate $y$:
$(3x + 2y) + (x - 2y) = 12 + 4$
$4x = 16$
$x = \frac{16}{4} = 4$

Substitute $x = 4$ into the second equation:
$4 - 2y = 4$
$-2y = 0$
$y = 0$

The solution is $(4, 0)$.

Here are the solutions so far:
* THE: $(3, -1)$
* STUFFED: $(9, 5)$
* ARE: $(4, 0)$