Solve the system of linear equations.

Step-by-step explanation:

Let's solve this by eliminating z, then we'll go from there.

Add 6 times the second equation to the first.

  (3x -3y +6z) +6(x +2y -z) = (3) +6(4)

  9x +9y = 27 . . . simplify

  x + y = 3 . . . . . . divide by 9 [eq4]

Add 13 times the second equation to the third.

  (5x -8y +13z) +13(x +2y -z) = (2) +13(4)

  18x +18y = 54

  x + y = 3 . . . . . . divide by 18 [eq5]

Equations [eq4] and [eq5] are identical. This tells us the system is dependent, and has an infinite number of solutions. We can find them in terms of z:

  y = 3 -x . . . . solve eq5 for y

  x +2(3 -x) -z = 4 . . . . substitute into the second equation

  -x +6 -z = 4

  x = 2 - z . . . . . . add x-4

  y = 3 -(2 -z)

  y = z +1

So far, we have written the solutions in terms of z. If we use the parameter "a", we can write the solutions as ...

  x = 2 -a

  y = 1 +a

  z = a

Check

First equation:

  3(2-a) -3(a+1) +6a = 3

  6 -3a -3a -3 +6a = 3 . . . true

Second equation:

  (2-a) +2(a+1) -a = 4

  2 -a +2a +2 -a = 4 . . . true

Third equation:

  5(2-a) -8(a+1) +13a = 2

  10 -5a -8a -8 +13a = 2 . . . true

Our solution checks algebraically.

step-by-step explanation: