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Question 915932: helppp
Jen butler has been pricing Speed-Pass train fares for a group trip to New York.
three adults and four children must pay $129.
two adults and three children must pay $92.
Find the price of the adults ticket and the price of a child's ticket.
Adult
Child
Answer by josh_jordan(263)   (Show Source):
You can put this solution on YOUR website! To solve, we need to set this word problem up as a system of linear equations. First, let's use the letter A to represent the Adult tickets and C to represent the child tickets. The problem says that 3 Adults and 4 children must pay 129 dollars. So,
3A + 4C = 129
Next, we are told that 2 Adults and 3 children must pay 92 dollars. So,
2A + 3C = 92
Our system of equations are
3A + 4C = 129
2A + 3C = 92
We need to now solve for either A or C. Let's choose C. To solve for C, we need to multiply the first equation by a certain number, so that when we add that first equation to the second equation, our letter A will disappear. If we multiply equation 1 by -2 and the second equation by 3 and then add them together, our A will go away:
-2(3A + 4C = 129) -----> -6A - 8C = -258
3(2A + 3C = 92) -----> 6A + 9C = 276
Now, let's add these two results together:
-6A - 8C + 6A + 9C = 276 - 258 -----> C = 18
Now that we have solved for C, we can replace C in either of our two original equations with 18. Let's use equation 1:
3A + 4C = 129 -----> 3A + 4(18) = 129 -----> 3A + 72 = 129
Next, subtract both sides by 72:
3A + 72 - 72 = 129 - 72 -----> 3A = 57
Finally, divide both sides by 3 to solve for A:
3A/3 = 57/3 -----> A = 19
Therefore, the price of an Adult ticket is $19 and the price of a Child ticket is $18.
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