Terry bought some gum and candy. The number of packages of chewing gum was one more than the number of mints. The number of mints was three times the number of candy bars. If gum was 6 cents a package, mints were 3 cents each, and candy bars were 10 cents each, how many of each did he get for 80 cents?
g = m+1
m = 3c so c= m/3
6g + 3m + 10c = 80
6(m+1) + 3m + 10m/3 = 80
6m + 6 + 3m + 10 m/3 = 80
9 m + 10m/3 = 74
37 m/3 = 74
m/3 = 2
m = 6
you can take it from there
I am stupid
To solve this problem, we need to set up equations based on the given information.
Let's assume that Terry bought x packages of gum, y mints, and z candy bars.
Given that the number of gum packages was one more than the number of mints, we can write the equation:
x = y + 1 ...(Equation 1)
Given that the number of mints was three times the number of candy bars, we can write the equation:
y = 3z ...(Equation 2)
Given that gum was 6 cents a package, mints were 3 cents each, and candy bars were 10 cents each, we can write the equation for the total cost:
6x + 3y + 10z = 80 ...(Equation 3)
Now, let's solve these equations to find the values of x, y, and z.
From Equation 2, we can substitute y in terms of z into Equation 1:
x = (3z) + 1
x = 3z + 1 ...(Equation 4)
Now, substitute the values of x and y in terms of z (from Equation 4 and 2, respectively) into Equation 3:
6(3z + 1) + 3(3z) + 10z = 80
18z + 6 + 9z + 10z = 80
37z + 6 = 80
37z = 80 - 6
37z = 74
z = 74 / 37
z = 2
Now, we can find the values of x and y using Equation 4 and Equation 2, respectively:
x = 3z + 1
x = 3(2) + 1
x = 6 + 1
x = 7
y = 3z
y = 3(2)
y = 6
Therefore, Terry bought 7 packages of gum, 6 mints, and 2 candy bars for a total of 80 cents.