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Here is the problem: negative x minus 2y equals 8; 2y minus z equals negative 4; 2x minus 3y minus 2z equals negative 2.

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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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Divide R1 by negative 1.

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Clear column 1 using the pivot in R1.

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Divide R2 by 2.

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Clear column 2 using the pivot in R2.

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Divide R3 by negative 11 over 2.

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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 4, y equals negative 2, z equals 0.

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

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Equation 1: negative x minus 2y equals 8, which checks out.

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Equation 2: 2y minus z equals negative 4, which checks out.

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