Select the correct answer. What is the value of \( b \) in this linear system? \( \begin{array}{r}a-b+c=-6 \\ b-c=5 \\ 2 a-2 c=4 \\ \text { A. }-3 \\ \text { B. } 2 \\ \text { C. } 1 \\ \text { D. }-1\end{array} \)

To solve for \( b \) in the given linear system, we can manipulate the equations step-by-step. Starting from the second equation: 1. From \( b - c = 5 \), we can express \( b \) as \( b = c + 5 \). Now, substitute \( b \) into the first equation: 2. Replace \( b \) in the first equation \( a - (c + 5) + c = -6 \): \[ a - c - 5 + c = -6 \] This simplifies to: \[ a - 5 = -6 \] Thus, \[ a = -6 + 5 = -1. \] Now that we have the value of \( a \), we can substitute it into the third equation to find \( c \): 3. The third equation is \( 2a - 2c = 4 \): \[ 2(-1) - 2c = 4 \] Simplifying gives: \[ -2 - 2c = 4. \] Rearranging this leads to: \[ -2c = 4 + 2 \Rightarrow -2c = 6 \Rightarrow c = -3. \] Finally, we can substitute \( c \) back into the expression for \( b \): 4. From \( b = c + 5 \): \[ b = -3 + 5 = 2. \] So, the value of \( b \) is \( \boxed{2} \).