Let x = number of children, y = number of students, and z = number of adults
x + y + z = 1200
2x + 3y + 4z = 3500
z = (1/3)(x + y)
(4/3)(x + y) = 1200
2x + 3y + 1200 = 3500
So, 2x + 3y = 2300
x + y = 900
Multiply the second equation by -2:
2x + 3y = 2300
-2x - 2y = -1800
Add the equations: y = 500
Back substitute to get x = 400, and z = 300

Bradford T.
The matrix by rows would be [1 1 1 1200], [2 3 4 3500], [1 1 -3 0] The last row comes from 3z=x+y or x+y-3z=004/23/24
Hannah J.
Thank you for your help, but for my college algebra we use a calculator and matrices, and I’m completely confused on what to input for z. I’m also still lost on how you found z.04/23/24