Solving Systems of Linear Equations by Elimination

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: \((4, 0)\)