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iren2701
2 months ago
8

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 num

ber 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?
Mathematics
2 answers:
Inessa [12.5K]2 months ago
8 0
2(2y-10)+3y=1240
7y-20=1240
7y=1260
y=180
2x+3(180)=1240
2x+540=1240
2x=700
x=350
Therefore, they sold 350 tickets for food and 180 tickets for rides.
#TeamAlvaxic
zzz [12.3K]2 months ago
8 0

Solution:

Here, x denotes the number of food tickets sold and y denotes the number of ride tickets sold.

Given the system of equations:

2x+3y = 1240         ....[1]

x = 2y-10              ....[2]

Replace [2] into [1] to obtain;

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

Apply the distributive property,

a \cdot(b+c) = a\cdot b+ a\cdot c

4y-20+3y = 1240

Combine like terms;

7y-20 = 1240

Add 20 to both sides to obtain;

7y= 1260

Divide both sides by 7 to get;

y = 180

Plug this value into [2] to find x;

x = 2(180)-10 = 360-10 = 350

Therefore, the number sold of each ticket type is:

the number of food tickets sold: 350 and the number of ride tickets sold: 180

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