Question 1203703
<pre>

B, C, and T must be non-negative integers:

{{{system(40B+20C+40T=500,10B+20C+70T=200)}}}

Subtract the 2nd eq. from the 1st.

{{{matrix(3,1,
30B-30T=300,
  B-T=10,
B=10+T)}}}

{{{system(40B+20C+40T=500,10B+20C+70T=200)}}}

Eliminate the constant, multiply 1st eq. by 2, and 2nd eq. by 5

{{{system(80B+40C+80T=1000,50B+100C+350T=1000)}}}

{{{matrix(15,1,
80B+40C+80T=50B+100C+350T,
30B=60C+270T,
B=2C+9T,
B-9T=2C,
C>=0,
B-9T>=0,
B>=9T,
B=10+T,
10+T>=9T,
10>=8T,
1.25>=T,
matrix(1,3,T=0,or,T=1),
matrix(1,3,B=10,or,B=11),
(B-9T)/2=C,
matrix(1,3,C=(10-9*0)/2=10/2=5,or,C=(11-9*1)/2=2/2=1)
)}}}

Answer: 10 buffets, 5 chairs, and no tables  OR
        11 buffets, 1 chair,  and  1 table.

Edwin</pre>