6x+8y= 100

x= 13-y

replace

2y+78= 100

2y=22

y= 1 1

x=2

The digital design of this topic is too coincidental. 13 dining table, all tables are full. 100 guests, everyone has a seat.

If it is 105 people, how can it be cost-effective? 107 people, how is it cost-effective? Of course, the 13 table is not enough. Minimum number of tables, each table should be as full as possible. That makes more sense.

This problem is too simple to work out with algebra.

In fact, it is better to exercise people's thinking with arithmetic than algebra.

For example, 13 table has 8 people, so 8X 13= 104 people. Change the table for eight people into a table for six, change it into one, and two people are missing. Just change two.

Realistic and simple: 100 people, 8 people at a table, 13 table. Finally, I found that two tables were not full, only six people were sitting, so I changed the people at these two tables to small tables.

In this way, if there are 105 people, there will be 14 tables, including eight immortals 13 tables and one table for six people. There were originally 13X8= 104 people, so it is most cost-effective to take one person from each of the five square tables to fill the tables of these six people.

There are other options. Replace two square tables with six tables, and then add the extra four to whose table, and it will be five people.

The result is 1 1 square table, three tables with six people. 1 1*8+2*6+5= 105

Which do you think is more reasonable?

The latter, of course. Because the standard of a table for 8 people is different from that of a table for 6 people. In the former scheme, five tables are not full, while in the latter scheme, only one table is not full.

105 people, 14 table questions, how to do it with algebra?

Or 95 people, 13 table, how to do it with algebra?

Why don't you try?