Devon and his friends bought strawberry wafers for $3 per packet and chocolate wafers for $1 per packet at a carnival. They spent a total of $30 to buy a total of 22 packets of wafers of the two varieties.

Complete question is;

Devon and his friends bought strawberry wafers for $3 per packet and chocolate wafers for $1 per packet at a carnival. They spent a total of $30 to buy a total of 22 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 Devon and his friends bought at the carnival. Define the variables used in the equations.

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.

A) System of equation is;

x + y = 22 - - - (eq 1)

3x + y = 30 - - - (eq 2)

B) They bought 4 packets of strawberry wafers and 18 packets of chocolate wafers.

- Elimination method was used to solve the system of equations because the coefficients of y in both equations were the same.

Step-by-step explanation:

A) Let x represent the number of strawberry wafer packets, and let y represent the number of chocolate wafer packets.

They bought a total of 22 packets, thus;

x + y = 22 - - - (eq 1)

Strawberry wafers costs $3 per packet and chocolate wafers costs $1 per packet. Since they spent a total of $30. Thus;

3x + y = 30 - - - (eq 2)

System of equation is;

x + y = 22 - - - (eq 1)

3x + y = 30 - - - (eq 2)

B) Let's use elimination method to get the numbers of each wager bought.

Subtract eq (1) from eq (2) to get;

3x - x + y - y = 30 - 22

2x = 8

x = 4

Put 4 for x in eq 1 to get;

4 + y = 22

y = 22 - 4

y = 18

Thus, they bought 4 packets of strawberry wafers and 18 packets of chocolate wafers.

Elimination method was used to solve the system of equations because the coefficients of y in both equations were the same.

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.

you would need to also lose 1 point for each of the other outcomes for it to be fair.

step-by-step explanation: