From the statement, we know that:
• the catering company expect 150 people,
,
• they provide two kinds of tables,
,
• small tables seat 6 people,
,
• large tables seat 10 people.
We define the variables
• x = # of small tables,
,
• y = # of large tables.
From the info above, we see that:
• 6x is the number of people that x small tables can sit,
,
• 10y is the number of people that y large tables can sit.
The sum of 6x and 10y gives us the total number that we can sit and that number must be equal to 150:
![6x+10y=150.](https://img.qammunity.org/2023/formulas/mathematics/college/m1ebb8qczlp4elak5u0und3ad6nyxkrdkw.png)
a) One possible combination is to choose:
• x = 10,
,
• y = 9.
We verify that the combination sum up 150 people
![6\cdot10+10\cdot9=60+90=150\text{ }✓](https://img.qammunity.org/2023/formulas/mathematics/college/n4e03yrgxna6agt63qug72ueqlrr8lfmgq.png)
b) We previously have obtained the equation that represents the relation between x and y, that equation is:
![6x+10y=150.](https://img.qammunity.org/2023/formulas/mathematics/college/m1ebb8qczlp4elak5u0und3ad6nyxkrdkw.png)
c) If we have the point (20, 5), we have:
• x = 20,
,
• y = 5.
We know that with 6 small tables and 10 larges tables we can sit:
![6x+10y\text{ people.}](https://img.qammunity.org/2023/formulas/mathematics/college/omld5wo5fxvnd8fs8jcyofq84pvyihj4pa.png)
By replacing x = 20 and y = 5 we get:
![6\cdot20+10\cdot5=120+50=170.](https://img.qammunity.org/2023/formulas/mathematics/college/g0qijar7eccmxdismcj7h8rolu2kxn9w3u.png)
This result means that we can sit 170 people with (x, y) = (20, 5).
d) The equation that we wrote for this problem is:
![6x+10x=150.](https://img.qammunity.org/2023/formulas/mathematics/college/sln724quzinsnqnrbgyp3ff8lj6da08hpi.png)
If we replace x = 20 and y = 5 we get:
![\begin{gathered} 6\cdot20+10\cdot5\questeq150, \\ 120+50\questeq150, \\ 170\\e150\text{ }✖ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fgb7n2vwnq3lin4w9b3q4t6jor3im8po20.png)
We see that by replacing (x, y) = (20, 5) we obtain different numbers on both sides of the equation. So we conclude that the point (x, y) = (20, 5) is not a solution to the equation.
Answers
a) A possible combination is to have 10 small tables and 9 large tables.
b) The relation between x and y is given by the equation:
![6x+10y=150](https://img.qammunity.org/2023/formulas/mathematics/college/wr5ju975afoogud97tj7xo4dsl50y5ebzy.png)
c) The point (x, y) = (20, 5) represents the situation where we can sit 170 people.
d) The point (x, y) = (20, 5) is not a solution to the equation.