Question

Solve the system by elimination. (Please explain how to do it)

Answer 1

Apply to Row 2 : Row 2 + Row 1

2x + 2y + 3z = 0

y + 4z = -3

2x + 3y + 3z = 5

Apply to Row 3: Row 3 - Row 1

2x + 2y + 3z = 0

y + 4z = -3

y = 5

Apply to Row 3: Row 3 - Row 2

2x + 2y + 3z = 0

y + 4z = -3

-4z = 8

Simplify rows

2x + 2y + 3z = 0

y + 4z = -3

z = -2

Note that the matrix is in echelon form now. The next steps are for back substitution.

Apply to Row 2: Row 2 - 4 Row 3

2x + 2y + 3z = 0

y = 5

z = -2

Apply to Row 1: Row 1 - 3 Row 3

2x + 2y = 6

y = 5

z = -2

Apply to Row 1: Row 1 - 2 Row 2

2x = -4

y = 5

z = 2

Simplify the rows

x = -2

y = 5

z = -2