Determine whether \( (0,2,1,-2) \) satisfies each equation in the following system of equations. Equation Satisfies? \[ \begin{array}{ll}5 x_{1}+2 x_{2}+2 x_{3}+x_{4}=5 & (1) \\ 6 x_{1}+3 x_{2}+2 x_{3}+x_{4}=5 & (2) \\ 6 x_{1}+3 x_{2}+3 x_{3}+x_{4}=6 & (3) \\ 4 x_{1}+2 x_{2}+2 x_{3}+x_{4}=5 & (4) \\ \text { Is }(0,2,1,-2) \text { a solution of the system of equations? } \\ \text { V Select } \\ \text { No } \\ \text { Yes }\end{array} \]

To determine whether the point \( (0,2,1,-2) \) satisfies each equation in the system, we’ll substitute \( x_1 = 0 \), \( x_2 = 2 \), \( x_3 = 1 \), and \( x_4 = -2 \) into each equation. 1. For the first equation \( 5 x_{1} + 2 x_{2} + 2 x_{3} + x_{4} = 5 \): \[ 5(0) + 2(2) + 2(1) + (-2) = 0 + 4 + 2 - 2 = 4 \quad \text{(Not satisfied: } 4 \neq 5\text{)} \] 2. For the second equation \( 6 x_{1} + 3 x_{2} + 2 x_{3} + x_{4} = 5 \): \[ 6(0) + 3(2) + 2(1) + (-2) = 0 + 6 + 2 - 2 = 6 \quad \text{(Not satisfied: } 6 \neq 5\text{)} \] 3. For the third equation \( 6 x_{1} + 3 x_{2} + 3 x_{3} + x_{4} = 6 \): \[ 6(0) + 3(2) + 3(1) + (-2) = 0 + 6 + 3 - 2 = 7 \quad \text{(Not satisfied: } 7 \neq 6\text{)} \] 4. For the fourth equation \( 4 x_{1} + 2 x_{2} + 2 x_{3} + x_{4} = 5 \): \[ 4(0) + 2(2) + 2(1) + (-2) = 0 + 4 + 2 - 2 = 4 \quad \text{(Not satisfied: } 4 \neq 5\text{)} \] Now compiling these results: - Equation (1): No - Equation (2): No - Equation (3): No - Equation (4): No Thus, \( (0,2,1,-2) \) does not satisfy any of the equations in the system. **Final Conclusion:** No, \( (0,2,1,-2) \) is not a solution of the system of equations.