Answer:
Choice b
(x = 1, y = 6, z = -6)
Step-by-step explanation:
The system of equations is
3x + 4y + 3z = 9 (1)
3x + 3y + 3z = 3 (2)
2x + 4y + 3z = 8 (3)
Subtract (2) from (1) to eliminate the 3x and 3y terms and solve for y
(1) - (2)
==> 3x + 4y + 3z - (3x + 3y +3z) = 9 - 3
==> 3x + 4y + 3z - 3x - 3y - 3z = 6
(3x-3x) + (4y-3y) + (3z-3z) = 6
4y - 3y = 6
y = 6
Subtract (3) from (1) to isolate the x term
(3x - 2x) + (4y-4y) + (3z-3z) = 9 -8
x = 1
Divide equation (2) on both sides by 32 to eliminate the coefficients
Equation (2) ÷ 3
==> (3x + 3y + 3z) / 2 = 3/3
==> x + y + z = 1 (4)
Substitute the values of x and y in Equation (4)
1 + 6 + z = 1
7 + z = 1
Subtract 7 from both sides
==> z = 1-7 = 6
Answer:
(x = 1, y = 6, z = -6)
