Question

((HELP QUICK PLEASE))

Clark and his friends bought strawberry wafers for $2 per packet and chocolate wafers for $3 per packet at a carnival. They spent a total of $30 to buy a total of 14 packets of wafers of the two varieties.

Part A: Write a system of equations that can be solved to find the number of packets of strawberry wafers and the number of packets of chocolate wafers that Clark and his friends bought at the carnival. Make sure you say what each variable means! (Which one is which wafer type?)
Part B: How many packets of chocolate wafers and strawberry wafers did they buy? Explain how you got the answer and why you selected a particular method to get the answer.​

Answer 1

Explanation:

A. S=strawberry, C=chocolate

S+C=14 equation 1

2s+3c=30 equation 2

B. -2(s+c)=-2(14) multiply equation 1 by -2 to use elimination method

-2s+-2c=-28 modified Equation 1

2s+ 3c=30. Equation 2

C=2 add above two equations.

solve for s

s+c=14

s+2=14

s=12

Used elimination method since it was easier to add the two equations.

check answer by inputting s,c values in either equation.

2s+3c=30

2(12)+3(2)=30

24+6=30

30=30

Answer 2

9514 1404 393

Answer:

• 2s+3c=30
• s+c=14
• 2 chocolate
• 12 strawberry

Explanation:

Part A:

I like to use variable names that remind me what they stand for. Here, we can use s and c for the numbers of strawberry and chocolate packets, respectively. The two equations we can write based on the problem statement are ...

2s + 3c = 30 . . . . . . . the total cost of the purchase

s + c = 14 . . . . . . . . . . the number of items purchased

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Part B:

Ordinarily, I would use one equation for the problem. That equation can be arrived at by substituting for s using the second equation. We choose to substitute for s so the only remaining variable is the one representing the highest-value contributor to the total cost.

s = 14 -c

2(14 -c) +3c = 30 . . . . substitute for s

28 +c = 30 . . . . . . . . . simplify (notice the coefficient of c is positive)

c = 2 . . . . . . . . . . . subtract 28

s = 14 -2 = 12 . . . . find s using the above equation

They bought 2 packets of chocolate and 12 packets of strawberry wafers.

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Explanation of choices

When the total purchase is n items and x of them are of one kind, then n-x is the number of the other kind. You don't need to write a second equation to figure that out. Hence the problem can be described by a single equation, such as the one highlighted above. Doing this lets you skip a variable definition and a couple of solution steps you don't really need.

The choice of variable as being the high-cost item ensures that its coefficient in the equation will be positive. Fewer errors are made when arithmetic is done with positive numbers. Again, recognizing that early on saves headaches down the line.

Getting to a generic solution

Note that the number we subtract (28) is the "would be" cost of all of the items purchased being the lowest-cost item (14 strawberry packets). This recognition can let you do the entire problem in your head. The difference between the actual cost and the "would be" cost is the result of replacing some number of the lowest-cost items with the higher-cost ones. The difference between total costs is divided by the difference in cost per item to find the number of the higher-cost item. That is the answer to all problems of this type. (30 -14(2))/(3-2) = 2 = the number of $3 packets.