Respuesta :
There is no question in Part A, but i do what i can.
Part A.
We know that:
n - number of packets of crackers
p - number of packets of pretzels
p = 2n+9
s - number of packets of peanuts
s = 3p = 3(2n+9) = 6n+27
So:
[tex]n+p+s=414\\\\n+(2n+9)+(6n+27)=414\\\\9n+36=414\quad|-36\\\\9n=378\quad|:9\\\\\boxed{n=42}\\\\\\ p=2n+9=2\cdot42+9=\boxed{93}\\\\s=6n+27=6\cdot42+27=\boxed{279}[/tex]
There are 42 packets of crackers, 93 packets of pretzels and 279 packets of peanuts.
Part B.
x - number of rows with exactly 2 passengers
y - number of rows with exactly 3 passengers
and:
[tex]\boxed{\begin{cases}x+y=48\\2x+3y=132\end{cases}}\\\\\\ \begin{cases}x+y=48\quad|\cdot2\\2x+3y=132\end{cases}\\\\\\ \begin{cases}2x+2y=96\\2x+3y=132\end{cases}\\\\--------(-)\\\\2x-2x+2y-3y=96-132\\\\-y=-36\\\\\boxed{y=36}\\\\\\x+y=48\\\\x+36=48\\\\x=48-36\\\\\boxed{x=12}[/tex]
There are 12 rows with exactly 2 passengers and 36 rows with exactly 3 passengers.
Part A.
We know that:
n - number of packets of crackers
p - number of packets of pretzels
p = 2n+9
s - number of packets of peanuts
s = 3p = 3(2n+9) = 6n+27
So:
[tex]n+p+s=414\\\\n+(2n+9)+(6n+27)=414\\\\9n+36=414\quad|-36\\\\9n=378\quad|:9\\\\\boxed{n=42}\\\\\\ p=2n+9=2\cdot42+9=\boxed{93}\\\\s=6n+27=6\cdot42+27=\boxed{279}[/tex]
There are 42 packets of crackers, 93 packets of pretzels and 279 packets of peanuts.
Part B.
x - number of rows with exactly 2 passengers
y - number of rows with exactly 3 passengers
and:
[tex]\boxed{\begin{cases}x+y=48\\2x+3y=132\end{cases}}\\\\\\ \begin{cases}x+y=48\quad|\cdot2\\2x+3y=132\end{cases}\\\\\\ \begin{cases}2x+2y=96\\2x+3y=132\end{cases}\\\\--------(-)\\\\2x-2x+2y-3y=96-132\\\\-y=-36\\\\\boxed{y=36}\\\\\\x+y=48\\\\x+36=48\\\\x=48-36\\\\\boxed{x=12}[/tex]
There are 12 rows with exactly 2 passengers and 36 rows with exactly 3 passengers.