Word problem for math homework

Question

Emilia Bought three adult tickets and one Childs ticket to a movie for a total of $32.00. Kylie bought two adult tickets and five children's tickets for a total of $49.50. Find the cost of one adult ticket and one student ticket.

Answer 1

Hi Chase,

To start we will treat the cost of a children's ticket and the cost of an adult's ticket as variables:

A = cost of 1 adult's ticket

C = cost of 1 children's ticket

We know that Emilia bought 3 adult tickets (3A) and 1 children's ticket (1C) for $32.00. We can write this in a different way as an equation:

equation 1:

3A + 1C = $32

We know that Kylie bought 2 adult tickets (2A) and 5 children's tickets (5C) for $49.50. We can also write this as an equation:

equation 2:

2A + 5C = $49.50

Now we have two equations which are both true:

Equation 1:

3A + 1C = $32

Equation 2:

2A + 5C = $49.50

and we can use substitution to solve for our variables. Lets look at the first equation again:

3A + 1C = $32

If we subtract 3A from both sides we get:

C = $32 - 3A

Since C is also a variable in equation 2, we can take a look at equation 2 again but write $32 - 3A instead of C. Now we get:

2A + 5*($32 - 3A) = $49.50

Multiply out the 5 to get:

2A + $160 - 15A = $49.50

Combine like terms to get:

$160 -13A = $49.50

Add 13A to both sides:

$160 = $49.50 + 13A

Subtract $49.95 from both sides:

$110.50 = 13A

Divide each side by 13:

$8.50 = A

So the cost of an adult ticket is $8.50 (Eight dollars and fifty cents)

Now that we know the value of A we can plug that into our equation for C:

C = $32 - 3A = $32 - 3*($8.50) = $6.50

So the cost of an adult ticket is $8.50 and the cost of a children's ticket is $6.50.

Answer 2

This is a typical systems of equations problem.

let a = the cost of and adult ticket and c= the cost of a children’s ticket.

3a+c=32. -15a - 5c = -160

2a+5c=49.5. 2a +5c =49.5

now multiply the top equation by -5 and add the two equations together to eliminate the c

so -13a = -110.5

so a = 8.50 And then plugging that into the top equation you will find that c = 6.50

Answer 3

First, we will define two variables: a - price of adult ticket, c - price of children ticket

Next, we will rewrite using a system of equations.

3a + c = 32

2a + 5c = 49.50

Next, we will rearrange the first equation and solve for c

3a + c = 32

c = 32 - 3a

Finally, plug in our new equation into the second equation, and solve for a.

2a + 5c = 49.50

2a + 5(32 - 3a) = 49.50

2a + 160 - 15a = 49.50

160 - 13a = 49.50

110.50 = 13a

a = $8.50

Plug in our value for a to find our value for c

3a + c = 32

3(8.50) + c = 32

25.50 + c = 32

c = $6.50

The price of one adult ticket is $8.50 and the price of one child ticket is $6.50

Answer 4

3a+1c=32.00

2a+5c=49.50

a=8.5

c=6.5

a=adult

c=child