Question

A child is playing games with empty soda cups. There are three sizes: small, medium, and large. After some experimentation

Answer 1

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