The fun guys game rental store charges an annual fee of $5 plus $5.50 per game rented. the game bank charges an annual fee of $35 plus $2.50 per game. for how many game rentals will the cost be the same at both stores? what is that cost?

10 games; $60
9 games; $55
4 games; $27
8 games; $49

10 games; $60

Step-by-step explanation:

Let x represent the number of games rented at either stores. Let y represent the cost of renting x games at Fun guys rental.

Let z represent the cost of renting x games at Games bank.

The Fun Guys game rental store charges an annual fee of $5 plus $5.50 per game rented.

The cost of x games will be

y = 5 + 5.5x

The Game Bank charges an annual fee of $35 plus $2.50 per game.

The cost of x games will be

z = 35 + 2.5x

Since we want to determine the number of games that will be played for the costs to be the same, we would equate y to z. It becomes

5 + 5.5x = 35 + 2.5x

5.5x - 2.5x = 35 - 5

3x = 30

x = 30/3 = 10

They will make 10 games rentals before the costs become the same.

To find the cost of 10 games, we will substitute x = 10 into one of the equations, it becomes

5 + 5.5× 10 = 5 + 55 = $60

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

43

step-by-step explanation: