Question

QUICK PLS 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?

180 food tickets and 293 ride tickets
180 food tickets and 350 ride tickets
293 food tickets and 180 ride tickets
350 food tickets and 180 ride tickets

Answer 1

No4- 350 food tickets and 180 ride tickets

2x+3y=1240

x=2y-10

2(2y-10)+3y=1240

4y-20+3y=1240

7y-20=1240

7y=1260

y=1260/7

y=180

putting y in the eq x=2y-10

x=(2×180)-10

= 360-10

= 350

x=350

y=180

Answer 2

The correct option is D) 350 food tickets and 180 ride tickets.

Step-by-step explanation:

Consider the provided information.

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

$2x + 3y = 1240$   ......(1)

$x = 2y - 10$    ......(2)

Substitute the value of x in equation 1.

$2(2y-10)+ 3y = 1240$

$4y-20+ 3y = 1240$

$7y = 1260$

$y=180$

Therefore, the number of ride tickets sold is 180.

Substitute the value of y in equation 2.

$x = 2(180) - 10\\x=350$

Thus, 350 tickets sold of food.

Hence, the correct option is D) 350 food tickets and 180 ride tickets.

#Learn more

Solve by substitution method​

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