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.
