Respuesta :
Answer:
Part A:
[tex]x+y=27 \\4x+y = 45[/tex]
Part B:
Number of vanilla wafers packets bought = 6 and
Number of chocolate wafers packets bought = 21
Step-by-step explanation:
Given:
Cost of each packet of vanilla wafer = $4
Cost of each packet of chocolate wafer = $1
Total number of packets bought = 27
Total money spent = $45
Part A:
To write a system of equations for the given situation.
Here, we do not know the number of packets bought for each type of wafer.
We just know that total number of packets bought and the total money spent in buying those packets.
Let us suppose,
Number of packets of vanilla wafers bought = [tex]x[/tex] and
Number of packets of chocolate wafers bought = [tex]y[/tex]
Total number of packets = Number of packets of vanilla wafers bought Plus Number of packets of chocolate wafers bought:
i.e. [tex]x+y=27 ...... (1)[/tex]
Money spent on vanilla wafers packets = [tex]4 \times x[/tex]
Money spent on chocolate wafers packets = [tex]1 \times y[/tex]
Total money spent = [tex]4x+y = 45 ...... (2)[/tex]
So, the system of equations is:
[tex]x+y=27 \\4x+y = 45[/tex]
Part B:
To find number of packets bought for each flavor = ?
Here, we have two equations and two variables [tex]x[/tex] and [tex]y[/tex].
As, the coefficients of [tex]y[/tex] in both the equations are same.
Let us subtract equation (1) from (2):
[tex]4x+y-x-y=45-27\\\Rightarrow 3x = 18\\\Rightarrow \bold{x =6}[/tex]
By equation (1), putting the value of [tex]x[/tex] and solving for [tex]y[/tex]:
[tex]6+y=27\\\Rightarrow \bold{y =21}[/tex]
So, number of vanilla wafers packets bought = 6 and
number of chocolate wafers packets bought = 21