Answer:
Between 150 and 450
Step-by-step explanation:
We are going to find the number by resolving a system of linear equations.
First we write the system equations :
[tex]C+S+A=750[/tex]
Where C : children, S : students and A : adults
The equation represents the ''full attendance''
The second equation :
[tex]3C+5S+7A=3450[/tex]
This equation represents the totaled receipts.
The system :
[tex]C+S+A=750\\3C+5S+7A=3450[/tex]
has the following associated matrix :
[tex]\left[\begin{array}{cccc}1&1&1&750\\3&5&7&3450\end{array}\right][/tex]
By performing elementary matrix operations we find that the matrix is equivalent to
[tex]\left[\begin{array}{cccc}1&0&-1&150\\0&1&2&600\\\end{array}\right][/tex]
The new system :
[tex]C-A=150\\S+2A=600[/tex]
Working with the equations :
[tex]C = 150 + A\\S = 600-2A[/tex]
Our solution vector is :
[tex]\left[\begin{array}{c}C&S&A\end{array}\right] =\left[\begin{array}{c}150+A&600-2A&A\end{array}\right][/tex]
For example :
If 0 adults attended ⇒ A = 0
[tex]C = 150 + 0 \\C = 150\\S = 600 - 2A\\S = 600[/tex]
This verify the totaled receipts equation :
150($3)+600($5) = $ 3450
A ≥ 0 ⇒ If A = 0 ⇒ C = 150
C = 150 is the minimum children attendance
From the equation :
[tex]S = 600 -2A[/tex]
S ≥0
600 - 2A ≥ 0
600 ≥ 2A
300≥ A
A is restricted to the interval [ 0, 300]
When A = 0 ⇒ C = 150
When A = 300 ⇒C = 150 + A = 150 + 300 = 450
Children ∈ [ 150,450]
With C being an integer number (including 0)
Also S and A are integer numbers (including 0)
