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:
[tex]6x+10y=150.[/tex]a) One possible combination is to choose:
• x = 10,
,• y = 9.
We verify that the combination sum up 150 people
[tex]6\cdot10+10\cdot9=60+90=150\text{ }✓[/tex]b) We previously have obtained the equation that represents the relation between x and y, that equation is:
[tex]6x+10y=150.[/tex]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:
[tex]6x+10y\text{ people.}[/tex]By replacing x = 20 and y = 5 we get:
[tex]6\cdot20+10\cdot5=120+50=170.[/tex]This result means that we can sit 170 people with (x, y) = (20, 5).
d) The equation that we wrote for this problem is:
[tex]6x+10x=150.[/tex]If we replace x = 20 and y = 5 we get:
[tex]\begin{gathered} 6\cdot20+10\cdot5\questeq150, \\ 120+50\questeq150, \\ 170\ne150\text{ }✖ \end{gathered}[/tex]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:
[tex]6x+10y=150[/tex]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.
