What is the solution to the following system?a(2, 3, 4)b(4, 3, –2)c(4, 3, 2)d(6, 7, –2)

The answer is x = 4, y = 3, z = 2.
Solution:
We solve this given system of equations using elimination. We choose to eliminate variable z first by taking the sum of the first and second equations:
     3x + 2y + z = 20 (equation 1)
     x - 4y - z = -10 (equation 2)
     (equation 1) + (equation 1) => 4x - 2y = 10 (equation 4)

We now select the second and third equations and eliminate the same variable z:
     x - 4y - z = -10 (equation 2)
     2x + y + 2z = 15 (equation 3)
     (equation2 multiply by 2) + (equation 3) => 2x - 8y - 2z = -20
                                                                           2x + y + 2z = 15
                                                                      => 4x - 7y = -5 (equation 5)

We solve for y by subtracting the fifth equation from the fourth:
     (equation 4) + (equation 5) => 5y = 15
                                                      y = 3

We substitute y = 3 into the fourth equation to find x.
     4x - 2y = 10
     4x - 2(3) = 10
     4x = 16
     x = 4

We use y = 3 and x = 4 and substitute into the first equation to solve for z.
     3x + 2y + z = 20
     3(4) + 2(3) + z = 20
     12 + 6 + z = 20
     z = 2
Therefore, the solution is x = 4, y = 3, z = 2.