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