Question

Bentley went into a bakery and bought 4 donuts and 10 cookies, costing a total of $23. Skylar went into the same bakery and bought 8 donuts and 6 cookies, costing a total of $25. Determine the price of each donut and the price of each cookie.

Answer 1

Donuts cost $2.00 and Cookies cost $1.50

Explanation:

D = cost of a donut

C = cost of a cookie

4D + 10C = $23.00

8D + 6C = $25.00

Eliminate a variable when subtracting the two equations. Change both values with C to 60 in order to eliminate the C variable and solve for D.

80D + 60C = $250.00 subtracted from 24D + 60C = $138

56D = $112.00 (Divide by 56 to single out the variable)

56D/56 = $112.00/56

D = $2.00

Use the D value to solve for C.

4(2) + 10C = $23.00

8 + 10C = $23.00

8 - 8 + 10C = $23.00 - 8

10C = $15.00

10C/10 = $15/10

C = $1.50

Check:

Bentley:

4D + 10C = $23

4(2) + 10(1.50) = $23

8 + 15 = $23

23 = 23

Skylar:

8D + 6C = $25

8(2) + 6(1.50) = $25

16 + 9 = $25

25 = 25

Answer 2

Each donut costs $2 and each cookie costs $1.5

Explanation:

1. Let´s name the variables as the following:

x = price of one donut

y = price of one cookie

2. Write in an equation form which Bentley bought:

`4x+10y=23` (Eq.1)

3. Write in an equation form which Skylar bought:

`8x+6y=25` (Eq.2)

4. Solve for x in Eq.1:

`4x+10y=23`

`4x=23-10y`

`x=(23-10y)/(4)` (Eq.3)

5. Replace Eq.3 in Eq.2 and solve for y:

`8*((23-10y)/(4))+6y=25`

`(184-80y)/(4)+6y=25`

`(184-80y+24y)/(4)=25`

`184-80y+24y=100`

`-80y+24y=100-184`

`-56y=-84`

`y=(84)/(56)`

`y=1.5`

6. Replacing the value of y in Eq.3:

`x=(23-10*(1.5))/(4)`

`x=(23-10*(1.5))/(4)`

`x=(23-15)/(4)`

`x=(8)/(4)`

`x=2`

Therefore each donut costs $2 and each cookie costs $1.5