A party rental company has chairs and tables for rent. The total cost to rent 2 chairs and 3 tables is $28. The total cost to rent 4 chairs and 8 tables is $73.

Question

A party rental company has chairs and tables for rent. The total cost to rent 2

chairs and 3

tables is $28

. The total cost to rent 4

chairs and 8

tables is $73

. What is the cost to rent each chair and each table?

Answer 1

This is essentially a problem involving a system of equations.

 

Let c be the cost of one chair and let t be the cost of one table.

 

Then we have 2c + 3t = 28

              and 4c + 8t = 73

 

Let's solve the first equation for c:

 

2c + 3t = 28

2c = 28 - 3t (subtract 3t from both sides)

c = 14 - 3t/2 (divide each side by 2)

 

Now we can plug this into our second equation:

4 (14 - 3t/2) + 8t = 73. Now, we only have one variable and can solve:

 

56 - 6t + 8t = 73 (Distribute the 4)

56 + 2t = 73 (combine like terms)

2t = 17 (subtract 56 from both sides)

t = 8.5 (divide each side by 2)

 

Then c = 14 - 3(8.5)/2 = 14 - 12.75 = 1.25

 

So one table costs $8.50 and one chair costs $1.25

Answer 2

Let C=chairs and T= tables.

 

 

2C + 3T =28    -2(2C + 3T) =-4C + -6T = -56   → multiply by -2 
4C + 8T = 73                        4C  +  8T =  73

                                                      2T  = 17→    divide by 2

                                                      T    =  8.50

 

so plug in, 2C + 3(8.50) =28→using 2C + 3T =28

                2C + 25.50   = 28

                2C  -25.50    = -25.50

                 2C =               2.50→divide by 2

                    C = 1.25

 

so the cost of one table is $8.50 and the cost for one chair $ 1.25