maraDubenjiv5eymu
contestada

At a carnival, food tickets cost $2 each and ride tickets cost $3 each. A total of $1,240 was collected at the carnival. The number of food tickets sold was 10 less than twice the number of ride tickets sold.

The system of equations represents x, the number of food tickets sold, and y, the number of ride tickets sold.

2x + 3y = 1240

x = 2y – 10

How many of each type of ticket were sold?

Respuesta :

taskmasters

The given equations satisfy the given conditions. There are 2 equations and 2 unknowns, so a certain solution can be found.

This can be solved using substitution,

Substituting eqn 2 to eqn 1:

2(2y – 10) + 3y =1240

Simplifying,

y = 180

x = 350

The number of each type of tickets sold are as follows:

number of food ticket sold = 350

number of ride ticket sold = 180

How to solve for the variable in a system of equation?

x = number of food sold

y = number of ride tickets sold

Therefore,

2x + 3y = 1240

x = 2y – 10

Hence,

2x + 3y = 1240

x - 2y  = – 10

2x + 3y = 1240

2x - 4y = -20

7y = 1260

y = 1260 / 7

y = 180

Therefore,

x = 2(180) - 10

x = 360 - 10

x = 350

Therefore, 350 food tickets were sold and 180 ride tickets were sold.

learn more on equation here: https://brainly.com/question/1396677

#SPJ5

Otras preguntas