Question

What is the solution to the following system?

3x+10y-12z=40
x-5y=0
x-4z=0
(8, 40, 32)
(10, 2, 3)
(20, 4, 5)
(40, 8, 10)

Answer 1

x = 5y and x = 4z, so 5y = 4z ⇒ z = (5/4)y. Substitute into 3x+10y−12z=40 with x=5y:
3(5y)+10y−12(5/4 y)=40 → 15y+10y−15y=40 → 10y=40 → y=4.
Thus x=5y=20, z=5. Solution: (20, 4, 5).