Find all solutions to the linear system:

  1. \(x + y - z = -1\)
  2. \(x - y - z = -1\)
  3. \(z = 2\)

Given that the variable \(z\) is already given, I will solve this problem simply by substituting the value for \(z\) into the previous equations, and solve for \(x\) and \(y\) using simultaneos equations.

Substitute equation 3 into equations 1 & 2:

  1. \(x + y - 2 = -1\)
  2. \(x - y - 2 = -1\)

Simplify equations 1 & 2:

  1. \(x + y = 1\)
  2. \(x - y = 1\)

There is an obvious solution to these equations but let’s solve them manually.

Make \(x\) the subject of equation 2:

\(x = y + 1\)

Substitute \(x\) into equation 1 and solve for \(y\):

\(y + 1 + y = 1\)

\(2y = 0\)

\(y = 0\)

Substitute \(y\) into equation 2 and solve for \(x\):

\(x + 0 = 1\)

\(x = 1\)

There is a single solution to the system of equations: \(x = 1\), \(y = 0\), \(z = 2\).