at a carnival, food tickets cost $2 each and ride tickets cost $3 each. a total of $1,240 was collected at…

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

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

Explanation:

Step1: Substitute x in first - equation

Substitute $x = 2y-10$ into $2x + 3y=1240$. We get $2(2y - 10)+3y=1240$.

Step2: Expand the left - hand side

Expand $2(2y - 10)$ to get $4y-20$. So the equation becomes $4y-20 + 3y=1240$.

Step3: Combine like terms

Combine $4y$ and $3y$: $7y-20 = 1240$.

Step4: Isolate the variable y

Add 20 to both sides: $7y=1240 + 20=1260$. Then divide both sides by 7, so $y=\frac{1260}{7}=180$.

Step5: Find the value of x

Substitute $y = 180$ into $x = 2y-10$. Then $x=2\times180-10=360 - 10=350$.

Answer:

D. 350 food tickets and 180 ride tickets