A party rental company has chairs and tables for rent. The total cost to rent 2 chairs and 3 tables is$31 . The total cost to rent 6 chairs and 5 tables is $59 . What is the cost to rent each chair and each table?

The cost to rent each chair is $2.75 and the cost to rent each table is $8.50

Step-by-step explanation:

Let the:

Cost to rent a chair = x

Cost to rent a table = y

We would form an algebraic equation.

The total cost to rent 2 chairs and 3 tables is $31

2x + 3y = 31 Equation 1

The total cost to rent 6 chairs and 5 tables is $59

6x + 5y = 59 Equation 2

We solve the above equation above using elimination method

Multiply Equation 1 all through by the coefficient of x = 6 in Equation 2

Multiply Equation 2 all through by the coefficient of x = 2 in Equation 1

Hence, we have:

2x + 3y = 31 Equation 1 × 6

6x + 5y = 59 Equation 2 × 2

12x + 18y = 186 Equation 3

12x + 10y = 118 .… Equation 4

Subtracting Equation 4 from Equation 3

= 8y = 68

y = 68/8

y = 8.5

Therefore, the cost to rent a table = $8.50

Substituting 8.5 for y in Equation 1 to get the value of x

2x + 3y = 31 Equation 1

2x + 3(8.5) = 31

2x = 31 - 3(8.5)

2x = 31 - 25.5

2x = 5.5

x = 5.5/2

x = 2.75

The cost to rent a chair = $2.75

Therefore, the cost to rent each chair is $2.75 and the cost to rent each table is $8.50

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point

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step-by-step explanation: