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.


Respuesta :

Answer:

Donuts cost $2.00 and Cookies cost $1.50

Step-by-step 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:

Each donut costs $2 and each cookie costs $1.5

Step-by-step 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:

[tex]4x+10y=23[/tex] (Eq.1)

3. Write in an equation form which Skylar bought:

[tex]8x+6y=25[/tex] (Eq.2)

4. Solve for x in Eq.1:

[tex]4x+10y=23[/tex]

[tex]4x=23-10y[/tex]

[tex]x=\frac{23-10y}{4}[/tex] (Eq.3)

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

[tex]8*(\frac{23-10y}{4})+6y=25[/tex]

[tex]\frac{184-80y}{4}+6y=25[/tex]

[tex]\frac{184-80y+24y}{4}=25[/tex]

[tex]184-80y+24y=100[/tex]

[tex]-80y+24y=100-184[/tex]

[tex]-56y=-84[/tex]

[tex]y=\frac{84}{56}[/tex]

[tex]y=1.5[/tex]

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

[tex]x=\frac{23-10*(1.5)}{4}[/tex]

[tex]x=\frac{23-10*(1.5)}{4}[/tex]

[tex]x=\frac{23-15}{4}[/tex]

[tex]x=\frac{8}{4}[/tex]

[tex]x=2[/tex]

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

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