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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 :

MartynBH
2(2y-10)+3y=1240
7y-20=1240
7y=1260
y=180
2x+3(180)=1240
2x+540=1240
2x=700
x=350
So they sold 350 food tickets and 180 ride tickets in total.
#TeamAlvaxic

Answer:

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

Given the system of equations:

[tex]2x+3y = 1240[/tex]         ....[1]

[tex]x = 2y-10[/tex]              ....[2]

Substitute the equation [2] into [1] we have;

[tex]2(2y-10)+3y = 1240[/tex]

Using distributive property,

[tex]a \cdot(b+c) = a\cdot b+ a\cdot c[/tex]

[tex]4y-20+3y = 1240[/tex]

Combine like terms;

[tex]7y-20 = 1240[/tex]

Add 20 to both sides we have;

[tex]7y= 1260[/tex]

Divide both sides by 7 we have;

y = 180

Substitute this in [2] we have;

[tex]x = 2(180)-10 = 360-10 = 350[/tex]

therefore, each type of ticket were sold are:

the number of food tickets sold is, 350 and the number of ride tickets sold is, 180

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