Let's use a system of equations to solve this problem:
3p + 2s = 8.75 (Equation 1)
2p + 4s = 8.50 (Equation 2)
We want to find the cost of 1p and 3s, so we can set up a third equation:
p + 3s = C (Equation 3)
We need to solve for C. We can do this by first solving for p in terms of s in Equation 2:
2p + 4s = 8.50
2p = 8.50 - 4s
p = 4.25 - 2s
Now we can substitute this expression for p into Equation 3:
4.25 - 2s + 3s = C
4.25 + s = C
So 1p and 3s would cost C = 4.25 + s. To find the value of s, we can substitute the values from Equation 1:
3p + 2s = 8.75
3(4.25 - 2s) + 2s = 8.75
12.75 - 6s + 2s = 8.75
-4s = -4
s = 1
Therefore, 1p and 3s would cost C = 4.25 + s = 4.25 + 1 = 5.25.