A party rental company has chairs and tables to rent. There were two customers who rented both chairs and tables last week. The table below shows the number of chairs, the number of tables, and the total cost (in dollars) for those two customers.
First customer SEcond customer
Number of 3 5
chairs
Number of 8 2
tables
Total cost 71 22
Let x be the cost (in dollars) to rent a chair.
Let y be the cost (in dollars) to rent a table.
Write a system of equations that could be used to find the rental cost (in dollars) of each chair and each table.
(in dollars)
(b) How much does each chair and each table cost (in dollars) to rent?
The system of equations that could be used to find the rental cost of each chair and each table is:
3x + 8y = 71
5x + 2y = 22
To solve for x and y, we can use elimination or substitution. In this case, let's use the elimination method.
Multiplying the first equation by 5 and the second equation by 3, we get:
15x + 40y = 355
15x + 6y = 66
By subtracting the second equation from the first, we can eliminate x:
15x + 40y - (15x + 6y) = 355 - 66
34y = 289
Dividing both sides by 34, we find:
y = 289/34
Now, substituting this value back into either of the original equations, let's use the first one:
3x + 8(289/34) = 71
Multiplying through by 34 to eliminate the fraction, we obtain:
102x + 2312 = 2414
Subtracting 2312 from both sides, we have:
102x = 102
Dividing both sides by 102, we find:
x = 1
So, each chair costs 1 dollar to rent and each table costs (289/34) dollars to rent.