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Here is the problem: - 2y minus 2z equals 2; 2x plus 2y minus 3z equals negative 14; 2y plus 3z equals 0.

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3 linear equations in 3 unknowns (x, y, z).

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Write every equation in standard form. Move all variable terms to the left and constants to the right.

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Build the augmented matrix [A | b]. Columns: x, y, z | constant.

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Gauss–Jordan elimination. Use row operations to make each pivot 1 and clear the rest of its column.

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Swap R1 and R2.

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Divide R1 by 2.

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Divide R2 by negative 2.

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Clear column 2 using the pivot in R2.

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Clear column 3 using the pivot in R3.

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Reduced row-echelon form.

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Read off the unique solution.

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So the answer is x equals negative 1, y equals negative 3, z equals 2.

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The system has exactly one solution; substituting it back satisfies every equation.

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Equation 1: negative 2y minus 2z equals 2, which checks out.

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Equation 2: 2x plus 2y minus 3z equals negative 14, which checks out.

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Equation 3: 2y plus 3z equals 0, which checks out.