The local ice cream parlor sells small, medium, and large dishes of ice cream. One family
purchased a small, medium, and large dish for $21. A second family purchased three small, one
medium, and one large dish for $31. A third family purchased one small, one medium, and two
large dishes for $31. Determine the price of each product size.
Let's assume the price of a small dish of ice cream is $x, the price of a medium dish is $y, and the price of a large dish is $z.
First, we can set up the following equations based on the information given:
1. For the first family: x + y + z = 21
2. For the second family: 3x + y + z = 31
3. For the third family: x + y + 2z = 31
To solve this system of equations, we can use substitution or elimination method. Let's use the elimination method.
Subtracting equation 1 from equation 2, we get:
(3x + y + z) - (x + y + z) = 31 - 21
2x = 10
x = 5
Now, substitute the value of x back into equation 1 or 2:
x + y + z = 21
5 + y + z = 21
y + z = 16
Next, substitute the values of x and y into equation 3:
x + y + 2z = 31
5 + 16 + 2z = 31
21 + 2z = 31
2z = 10
z = 5
Therefore, the price of a small dish is $5, the price of a medium dish is $16, and the price of a large dish is $5.
Let's denote the price of a small dish as $s, the price of a medium dish as $m, and the price of a large dish as $l.
From the information given, we can set up the following equations:
Equation 1: s + m + l = 21 (family 1's purchase)
Equation 2: 3s + m + l = 31 (family 2's purchase)
Equation 3: s + m + 2l = 31 (family 3's purchase)
We can solve this system of equations using the substitution method or elimination method. Let's use the elimination method:
Multiply Equation 1 by 3 to match the coefficient of s in Equation 2:
3(s + m + l) = 3(21)
3s + 3m + 3l = 63 (equation 1 multiplied by 3)
Subtract Equation 1 from Equation 2:
(3s + m + l) - (s + m + l) = 31 - 21
3s - s + m - m + l - l = 10
2s = 10
s = 5
Substitute the value of s into Equation 1:
5 + m + l = 21
m + l = 16 (equation 4)
Substitute the value of s into Equation 3:
5 + m + 2l = 31
m + 2l = 26 (equation 5)
Now we have two equations with two variables (equations 4 and 5). We can solve this system of equations.
Multiply equation 4 by 2:
2(m + l) = 2(16)
2m + 2l = 32
Subtract equation 5 from this new equation:
(2m + 2l) - (m + 2l) = 32 - 26
m = 6
Substitute the value of m into equation 4:
6 + l = 16
l = 10
So, the price of a small dish is $5, the price of a medium dish is $6, and the price of a large dish is $10.
To solve this problem, we need to set up a system of equations based on the information given. Let's represent the price of a small dish as 's', the price of a medium dish as 'm', and the price of a large dish as 'l'.
From the first family:
1s + 1m + 1l = 21 (Equation 1)
From the second family:
3s + 1m + 1l = 31 (Equation 2)
From the third family:
1s + 1m + 2l = 31 (Equation 3)
We now have a system of three equations with three unknowns. To solve this system, we can use a method called substitution or elimination.
One way to solve this system is by using the substitution method. Here's how it works:
Step 1: Solve Equation 1 for 's':
s = 21 - m - l
Step 2: Substitute the value of 's' from Step 1 into Equations 2 and 3:
3(21 - m - l) + 1m + 1l = 31 (Equation 2)
1(21 - m - l) + 1m + 2l = 31 (Equation 3)
Step 3: Simplify and solve the resulting system of equations. Let's solve Equation 2 first:
63 - 3m - 3l + m + l = 31
63 - 2m - 2l = 31
-2m - 2l = 31 - 63
-2m - 2l = -32
Step 4: Now let's solve Equation 3:
21 - m - l + m + 2l = 31
21 + l = 31
l = 31 - 21
l = 10
Step 5: Substitute the value of 'l' into Equation 1 to solve for 's':
1s + 1m + 10 = 21
s + m = 21 - 10
s + m = 11
s = 11 - m
Step 6: Substitute the values of 's' and 'l' into Equation 1 or Equation 2 to solve for 'm':
s + m = 11
(11 - m) + m = 11
11 - m + m = 11
11 = 11
This means that 'm' can have any value, as long as 's' and 'l' are determined as specified.
So, the price of a small dish (s), medium dish (m), and large dish (l) cannot be determined with the given information.