Jen butler has been pricing Speed Pass train fares for a group trip to New York. Three adults and four children must pay $97. Two adults and three children must pay $69. Find the price of the adult ticket and price of a child’s ticket.
3 a + 4 c = 97
2 a + 3 c = 69
6 a + 8 c = 194
6 a + 9 c = 207
--------------------- subtract
- c = - 113
c = 113
go back for a
whoops typo, c= 13
Let's denote the price of an adult ticket as "a" and the price of a child's ticket as "c".
From the given information, we can set up two equations:
Equation 1: 3a + 4c = 97
Equation 2: 2a + 3c = 69
We can solve this system of equations using substitution or elimination method. Let's use the elimination method here.
Multiply Equation 2 by 2:
2(2a + 3c) = 2(69)
4a + 6c = 138
Now, we can subtract Equation 1 from this new equation to eliminate "a":
(4a + 6c) - (3a + 4c) = 138 - 97
4a + 6c - 3a - 4c = 41
Simplifying, we get:
a + 2c = 41
Now, we have a new equation with just "a" and "c".
Next, we can multiply Equation 1 by 2:
2(3a + 4c) = 2(97)
6a + 8c = 194
Now, we can subtract Equation 2 from this new equation to eliminate "a":
(6a + 8c) - (2a + 3c) = 194 - 69
6a + 8c - 2a - 3c = 125
Simplifying, we get:
4a + 5c = 125
We now have two equations:
a + 2c = 41
4a + 5c = 125
Now we can solve this system of equations to find the values of "a" and "c".
Multiplying Equation 1 by 4:
4(a + 2c) = 4(41)
4a + 8c = 164
Subtracting this new equation from Equation 2:
(4a + 5c) - (4a + 8c) = 125 - 164
4a + 5c - 4a - 8c = -39
Simplifying, we get:
-3c = -39
Dividing both sides of the equation by -3:
c = (-39) / (-3)
c = 13
Now, substitute the value of "c" back into Equation 1:
a + 2(13) = 41
a + 26 = 41
Subtracting 26 from both sides:
a = 41 - 26
a = 15
Therefore, the price of an adult ticket is $15 and the price of a child's ticket is $13.
To find the price of the adult ticket and the price of a child's ticket, let's set up an equation using the given information.
Let's assume the price of an adult ticket is "A" and the price of a child's ticket is "C".
According to the information given, we can create two equations:
Equation 1: 3A + 4C = 97 (Three adults and four children must pay $97)
Equation 2: 2A + 3C = 69 (Two adults and three children must pay $69)
To solve this system of equations, we can use the method of substitution:
Step 1: Solve Equation 2 for "A" in terms of "C":
2A + 3C = 69
2A = 69 - 3C (Subtract 3C from both sides)
A = (69 - 3C)/2 (Divide both sides by 2)
Step 2: Substitute the value of "A" from Step 1 into Equation 1:
3((69 - 3C)/2) + 4C = 97
Now, let's simplify and solve for "C":
(207 - 9C)/2 + 4C = 97 (Distribute 3 to (69 - 3C))
207 - 9C + 8C = 194 (Multiply both sides by 2)
-C = -13 (Combine like terms)
C = 13 (Divide both sides by -1)
Therefore, the price of a child's ticket, "C", is $13.
Step 3: Substitute the value of "C" into Equation 2 to find the price of the adult ticket, "A":
2A + 3(13) = 69
2A + 39 = 69 (Multiply 3 by 13)
2A = 69 - 39 (Subtract 39 from both sides)
2A = 30 (Combine like terms)
A = 15 (Divide both sides by 2)
Therefore, the price of an adult ticket, "A", is $15.
The price of an adult ticket is $15, and the price of a child's ticket is $13.