Exercise SSLE.C31
Find all solutions to the linear system:

  1. \(3x + 2y = 1\)
  2. \(x − y = 2\)
  3. \(4x + 2y =2\)

Solve equation 2 for x:
\(x = y + 2\)

Plug new equation 2 into other two equations:
\(3(y + 2) + 2y = 1\)
\(4(y+2) + 2y = 2\)

Simplify:
\(3y + 6 + 2y = 1\)
\(4y + 8 + 2y = 2\)

Combine like terms:
\(5y = -5\)
\(6y = -6\)

Solve for y:
\(y = -1\)
\(y = -1\)

Plug in y=-1 to solve for x:
\(x = -1 + 2 = 1\)

Check solution:
\(3(1) + 2(-1) = 1\)
\(3-2 = 1\)
\(1 = 1\)

\(1 - (-1) = 2\)
\(2 = 2\)

\(4(1) + 2(-1) = 2\)
\(4 - 2 = 2\)
\(2 = 2\)

The solution solves the system. Therefore, \(x = 1\) and \(y = -1\)