Answer:
S is the volume of the small cup, M the volume of the medium cup and L the volume of the large cup:
2*S + 2*M + 4*L = 160oz
2*S + 6*M + 1*L = 160oz
5*S + 1*M + 3*L = 160oz.
First, we must isolate one of the variables, for this we can use the first two equations and get:
2*S + 2*M + 4*L = 160oz = 2*S + 6*M + 1*L
We can cancel 2*S in both sides:
2*M + 4*L = 6*M + 1*L
now each side must have only one variable:
4*L - 1*L = 6*M - 2*M
3*L = 4*M
L = (4/3)*M.
now we can replace it in the equations and get :
2*S + 2*M + 4*(4/3)*M = 160oz
2*S + 6*M + (4/3)*M = 160oz
5*S + 1*M + 4M = 160oz.
simplifing them we have:
2*S + (22/3)*M + = 160oz
2*S + (22/3)*M = 160oz
5*S + 5*M = 160oz.
(the first and second equation are equal because we used those to get the relation of M and L, so we now have only two equations):
2*S + (22/3)*M = 160oz
5*S + 5*M = 160oz.
We can take the second equation and simplify it:
S + M = 160oz/5 = 32oz
S = 32oz - M
Now we can replace it in the first equation and solve it for M
2*S + (22/3)*M = 2*(32oz - M) + (22/3)*M = 160oz
62oz - 2*M + (22/3)*M = 160oz
-(6/3)*M + (22/3)*M = 98oz
(18/3)*M = 98oz
M = (3/18)*98oz = 16.33 oz
Then:
L = (4/3)*M =(4/3)*16.33oz = 21.78 oz
and:
S = 32oz - M = 32oz - 16.33oz = 15.67oz
