Question

Solve the system using the elimination method.

Answer 1

Answer:

  (x, y, z) = (-22/13, 29/13, 6/13)

Step-by-step explanation:

Adding the first and second equations, we get ...

  (3x +2y -3z) +(7x -2y +5z) = (-2) +(-14)

  10x +2z = -16 . . . . collect terms

  5x + z = -8 . . . . . . divide by 2 . . . [eq4]

Adding twice the second equation to the third, we get ...

  2(7x -2y +5z) +(2x +4y +z) = 2(-14) +(6)

  16x +11z = -22 . . . . . . [eq5]

Now, we have two equations with the variable y eliminated. We can subtract [eq5] from 11 times [eq4] to eliminate z:

  11(5x +z) -(16x +11z) = 11(-8) -(-22)

  39x = -66

  x = -66/39 = -22/13

From [eq4], we can find z as ...

  z = -8 -5x = -8 -5(-22/13) = 6/13

And from the second equation, we get ...

  y = (1/2)(7x +5z +14) = (1/2)(7(-22/13) +5(6/13) +14) = 29/13

The solution is (x, y, z) = (-22/13, 29/13, 6/13).