If 2 adult tickets and 1 child ticket cost $8 and if 1 adult ticket and 3 child tickets cost $9, what is the price of each ticket
2A + C = 8
A + 3C = 9
Multiply second equation by 2.
2A + 6C = 18
Subtract first equation from revised second.
5C = 10
Work it from there.
2a + c = 8
a + 3c = 9
I suggest substitution, from #1, let c = 8-2a
plug into #2
Well, it seems like the ticket prices are playing hide and seek. Let's solve this baffling mystery!
Let's call the price of an adult ticket "A," and the price of a child ticket "C."
From the first statement, we know that:
2A + C = 8
And from the second statement, we know:
A + 3C = 9
Now, let's put our detective caps on and solve this case!
To eliminate the "C" variable, we'll multiply the second equation by 2:
2A + 6C = 18
Now, let's subtract the first equation from the new equation:
(2A + 6C) - (2A + C) = 18 - 8
Simplifying, we get:
5C = 10
Therefore, the price of each child ticket, C, is $2.
Now, let's substitute the value of C into the first equation to find the price of an adult ticket:
2A + 2 = 8
2A = 6
A = 3
So, the price of each adult ticket, A, is $3.
To sum it up, each adult ticket costs $3, and each child ticket costs $2. Mystery solved!
To find the price of each ticket, we can set up a system of equations. Let's call the price of an adult ticket "a" and the price of a child ticket "c". Here are the equations that represent the given information:
Equation 1: 2a + 1c = 8
Equation 2: 1a + 3c = 9
To solve this system of equations, we can use the method of substitution or elimination. Let's use the method of substitution:
From Equation 1, we can express "c" in terms of "a" by rearranging the equation:
1c = 8 - 2a
c = 8 - 2a
Now, substitute this expression for "c" into Equation 2:
1a + 3(8 - 2a) = 9
1a + 24 - 6a = 9
-5a + 24 = 9
Next, isolate "a" by subtracting 24 from both sides:
-5a = 9 - 24
-5a = -15
Divide both sides by -5 to solve for "a":
a = -15 / -5
a = 3
Now that we have the value for "a", we can substitute it back into Equation 1 to find "c":
2(3) + 1c = 8
6 + 1c = 8
1c = 8 - 6
1c = 2
Therefore, the price of each adult ticket is $3 and the price of each child ticket is $2.
To find the price of each ticket, let's assign variables:
Let's say the price of an adult ticket is A, and the price of a child ticket is C.
From the given information, we can form two equations:
Equation 1: 2A + 1C = 8 (equation for 2 adult tickets and 1 child ticket)
Equation 2: 1A + 3C = 9 (equation for 1 adult ticket and 3 child tickets)
To solve this system of equations, we can use the method of substitution.
Let's solve Equation 1 for C:
C = 8 - 2A
Now substitute this value of C in Equation 2:
1A + 3(8 - 2A) = 9
1A + 24 - 6A = 9
-5A = 9 - 24
-5A = -15
A = -15 / -5
A = 3
Now substitute the value of A back into Equation 1 to find the value of C:
2(3) + 1C = 8
6 + 1C = 8
1C = 8 - 6
1C = 2
So, the price of an adult ticket (A) is $3, and the price of a child ticket (C) is $2.