Question 1187739: A party rental company has chairs and tables for rent. The total cost to rent 5 chairs and 2 tables is $18. The total cost to rent 3 chairs and 8 tables is $55. What is the cost to rent each chair and each table?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Let c be the rental cost for each chair
Let t be the rental cost for each table
The total cost to rent 5 chairs and 2 tables is $18: 5c + 2t = 18
The total cost to rent 3 chairs and 8 tables is $55. 3c + 8t = 55
(1) Solving the problem informally, using logical reasoning....
Since the cost of 5 chairs and 2 tables is $18, the cost of 20 chairs and 8 tables is 4*$18 = $72.
But the cost of 3 chairs and 8 tables is $55; the difference between the two cases is 20-3 = 17 more chairs for an additional $72-$55 = $17. So the rental cost for each chair is $17/17 = $1.
And then since 5 chairs at $1 each and 2 tables cost $18 to rent, the cost for the 2 tables is $13, so the cost per table is $6.50.
ANSWER: $1 per chair and $6.50 per table
CHECK:
5c+2t = 5(1)+2(6.50) = 5+13 = 18
3c+8t = 3(1)+8(6.50) = 3+52 = 55
(2) Using formal algebra....
5c + 2t = 18
3c + 8t = 55
Multiply the first equation by 4:
20c + 8t = 72
Compare that equation to the second original equation:
17c = 17
Solve to find the cost for each chair:
c = 17/17 = 1
Use that cost per chair in the first original equation to find the cost per table:
5(1)+2t = 18
5+2t = 18
2t = 18-5 = 13
t = 13/2 = 6.5
ANSWERS: c=1; t=6.5 -- i.e. $1 per chair and $6.50 per table
Observe that the solution using formal algebra uses EXACTLY the same calculations as the informal solution.
But we need to understand the formal algebraic solution, because an informal solution will not be possible when the problems get a lot more complicated.
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