Answer:
Explanation:
We can set up two equations: one for each sister:
Let C and D be the prices for a Candy bag and and Drink.
Sister 1: 3C + 1D = 3.20 [The total spent by sister 1 is the sum of 3 candy bags(C) and 1 drink (D), which comes to $3.20.]
Sister 2: 2C + 4D = 6.30 [The total spent by sister 2 is the sum of 2 candy bags(C) and 4 drinks (D), which comes to $6.30.]
We have two unknowns, C and D, BUT we have 2 equations. Therefore, we're likely going to be able to solve for C and D if we rearrange one of the equations to isolate one of the unknowns, and then use that expression in the other equation. Seems unlikely, but it always works.
I'll pick Sister 1's equation and solve for D, the price of a drink:
3C + 1D = 3.20
1D = 3.20 - 3C [This tells us the price of a drink is 3.20 - 3C]
Use this definition of 1D in Sister 2's equation:
2C + 4D = 6.30
2C + 4(3.20 - 3C) = 6.30
2C + 12.80 - 12C = 6.30
-10C = - 6.5
C = $0.65
One candy bag, C, is $0.65.
Use this value of C in any equation to find D, the cost of 1 drink. I'll pick The expression we derived above:
1D = 3.20 - 3C
1D = 3.20 - 3(0.65)
1D = 3.20 - 1.95
1D = $1.25
One drink, D, is $1.25.
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Check the values:
Candy Bags = $0.65 each Drinks = $1.25 each
Reported
Sister 1 2 1 2
Candy bags 3 2 $1.95 $1.30
Drinks 1 4 $1.25 $5.00
$3.20 $6.30 Calculated
$3.20 $6.30 Reported
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