Using the method of substitution, fill in the bla The solution set of the system \( \left\{\begin{array}{l}x-y-z=4 \\ x+y-z=12 \\ x-y+z=-4\end{array}\right. \)

To solve the system using substitution, let's start with one of the equations and express one variable in terms of the others. For simplicity, we can take the first equation \(x - y - z = 4\) and isolate \(x\): \[ x = y + z + 4 \] Now, we can substitute this expression for \(x\) into the other two equations. Substituting into the second equation \(x + y - z = 12\): \[ (y + z + 4) + y - z = 12 \] Simplifying this gives: \[ 2y + 4 = 12 \] \[ 2y = 8 \] \[ y = 4 \] Now that we have \(y\), we can substitute \(y = 4\) back into the expression for \(x\): \[ x = 4 + z + 4 \] So, \[ x = z + 8 \] Next, substitute \(y = 4\) into the first equation to find \(z\): \[ x - 4 - z = 4 \] Substituting \(x = z + 8\): \[ (z + 8) - 4 - z = 4 \] This simplifies to: \[ 4 = 4 \] This means we have found the relationship between \(x\), \(y\), and \(z\) but not specific values for \(z\). Hence, we can express \(x\) in terms of \(z\): Now let’s use the third equation \(x - y + z = -4\): Substituting \(y = 4\) and \(x = z + 8\): \[ (z + 8) - 4 + z = -4 \] This leads to: \[ 2z + 4 = -4 \] Simplifying gives: \[ 2z = -8 \] So, \[ z = -4 \] Now we can find \(x\) using \(z = -4\): \[ x = -4 + 8 = 4 \] We already found \(y = 4\). The solution set of the system is: \[ (x, y, z) = (4, 4, -4) \]