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.

Let x represents the number of food tickets sold and y represents  the number of ride tickets sold:

2x + 3y = 1240

x = 2y – 10

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

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

x = 360-10

x = 350

Simplifying,

y = 180

x = 350

so the number of ride tickets sold were 180 and number of food tickets were 350.

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