MATH SOLVE

4 months ago

Q:
# Jen Butler has been pricing Speed-Pass train fares for a group trip to New York. Three adults and four children must pay $ 104. Two adults and three children must pay $ 73. Find the price of the adult's ticket and the price of a child's ticket.

Accepted Solution

A:

An adult ticket is $20 and a child ticket is $11.

Our system of equations would be:

3A + 4C = 104

2A + 3C = 73

We want the coefficients of one of these variables to be the same in order to eliminate it; we will multiply the top equation by 2 and the bottom equation by 3:

2(3A + 4C = 104)

3(2A + 3C = 73)

6A + 8C = 208

6A + 9C = 219

Subtract the bottom equation:

6A + 8C = 208

-(6A + 9C = 219)

-1C = -11

Divide both sides by -1:

-1C/-1 = -11/-1

C = 11

Substitute this into the first equation:

3A + 4(11) = 104

3A + 44 = 104

Subtract 44 from both sides:

3A + 44 - 44 = 104 - 44

3A = 60

Divide both sides by 3:

3A/3 = 60/3

A = 20

Our system of equations would be:

3A + 4C = 104

2A + 3C = 73

We want the coefficients of one of these variables to be the same in order to eliminate it; we will multiply the top equation by 2 and the bottom equation by 3:

2(3A + 4C = 104)

3(2A + 3C = 73)

6A + 8C = 208

6A + 9C = 219

Subtract the bottom equation:

6A + 8C = 208

-(6A + 9C = 219)

-1C = -11

Divide both sides by -1:

-1C/-1 = -11/-1

C = 11

Substitute this into the first equation:

3A + 4(11) = 104

3A + 44 = 104

Subtract 44 from both sides:

3A + 44 - 44 = 104 - 44

3A = 60

Divide both sides by 3:

3A/3 = 60/3

A = 20