At a carnival, food tickets cost $2 each and ride 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.

Question:

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.

How many of each type of ticket were sold?

350 food tickets and 180 ride tickets were sold

Solution:

Let "x" be the number of food tickets sold

Let "y" be the number of ride tickets sold

Cost of 1 food ticket = $ 2

Cost of 1 ride ticket = $ 3

The number of food tickets sold was 10 less than twice the number of ride tickets sold

Number of food tickets sold = twice the number of ride tickets sold - 10

x = 2y - 10 -------- eqn 1

A total of $1,240 was collected at the carnival

Therefore, we frame a equation as:

number of food tickets sold x Cost of 1 food ticket + number of ride tickets sold x Cost of 1 ride ticket = 1240

2x + 3y = 1240 ------- eqn 2

Let us solve eqn 1 and eqn 2

Substitute eqn 1 in eqn 2

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

4y - 20 + 3y = 1240

7y = 1260

Substitute y = 180 in eqn 1

x = 2(180) - 10

x = 360 - 10

Thus 350 food tickets and 180 ride tickets were sold

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.

alright well the answers to your question is

2xy/9 hope this have a nice day : )

step-by-step explanation: