Solve the System of equation \[ \begin{array}{c}2 A+2 B+D=0 \\ -2 B+2 C-D+E=0 \\ 5 A+3 B-2 C+D-E=0 \\ -3 B+3 C-D+E=1 \\ 3 A-3 C-E=-3\end{array} \]

To solve the given system of equations, we will first rewrite them for clarity: 1. \( 2A + 2B + D = 0 \) (Equation 1) 2. \( -2B + 2C - D + E = 0 \) (Equation 2) 3. \( 5A + 3B - 2C + D - E = 0 \) (Equation 3) 4. \( -3B + 3C - D + E = 1 \) (Equation 4) 5. \( 3A - 3C - E = -3 \) (Equation 5) Now, let's solve the equations step by step. Start with Equation 1: \[ D = -2A - 2B \] Substituting \( D \) in Equations 2, 3, 4, and 5. Substituting into Equation 2: \[ -2B + 2C - (-2A - 2B) + E = 0 \] Simplifying yields: \[ 2A + 2C + E = 0 \quad \text{(Equation 6)} \] Substituting into Equation 3: \[ 5A + 3B - 2C + (-2A - 2B) - E = 0 \] This simplifies to: \[ 3A + B - 2C - E = 0 \quad \text{(Equation 7)} \] Substituting into Equation 4: \[ -3B + 3C - (-2A - 2B) + E = 1 \] Simplifying gives: \[ 2A - B + 3C + E = 1 \quad \text{(Equation 8)} \] Substituting into Equation 5: \[ 3A - 3C - E = -3 \] This stays as is. Now we have the following reduced system: 1. \( 2A + 2C + E = 0 \) (Equation 6) 2. \( 3A + B - 2C - E = 0 \) (Equation 7) 3. \( 2A - B + 3C + E = 1 \) (Equation 8) 4. \( 3A - 3C - E = -3 \) (Equation 5) We can express \( E \) from Equation 6: \[ E = -2A - 2C \] Substitute \( E \) into the other equations. Using \( E \) in Equation 7: \[ 3A + B - 2C - (-2A - 2C) = 0 \] Simplifying gives: \[ 5A + B = 0 \quad \text{or} \quad B = -5A \quad \text{(Equation 9)} \] Using \( E \) in Equation 8: \[ 2A - (-5A) + 3C - 2A - 2C = 1 \] This simplifies to: \[ 3A + C = 1 \quad \text{(Equation 10)} \] Now we need to use Equation 10. Express \( C \): \[ C = 1 - 3A \quad \text{(Equation 11)} \] Substituting Equation 11 into Equation 9: \[ B = -5A \] Now we substitute \( B \) and \( C \) into \( E \): \[ E = -2A - 2(1 - 3A) \] This simplifies to: \[ E = -2A - 2 + 6A = 4A - 2 \] Now we have: - \( A = A \) - \( B = -5A \) - \( C = 1 - 3A \) - \( D = -2A - 2B = -2A + 10A = 8A \) - \( E = 4A - 2 \) To find a particular solution, we can choose \( A = 0 \): - \( A = 0 \) - \( B = 0 \) - \( C = 1 \) - \( D = 0 \) - \( E = -2 \) Thus, one solution to the system is: \[ \boxed{(A, B, C, D, E) = (0, 0, 1, 0, -2)} \]