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Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineStanley loves riding Ferris wheels and roller coasters. While visiting the Harrison County Fair, he first went on the Ferris wheel 55 times and the roller coaster 22 times, using a total of 2020 tickets. Then, after taking a break and having a snack, Stanley went on the Ferris wheel 22 times and the roller coaster 55 times, using a total of 2929 tickets. How many tickets does it take to ride each attraction?\newlineIt takes _\_ tickets to ride the Ferris wheel, and _\_ tickets to ride the roller coaster.

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Q. Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineStanley loves riding Ferris wheels and roller coasters. While visiting the Harrison County Fair, he first went on the Ferris wheel 55 times and the roller coaster 22 times, using a total of 2020 tickets. Then, after taking a break and having a snack, Stanley went on the Ferris wheel 22 times and the roller coaster 55 times, using a total of 2929 tickets. How many tickets does it take to ride each attraction?\newlineIt takes _\_ tickets to ride the Ferris wheel, and _\_ tickets to ride the roller coaster.
  1. Define Variables: Let xx be the number of tickets needed for one ride on the Ferris wheel, and yy be the number of tickets needed for one ride on the roller coaster. Stanley first went on the Ferris wheel 55 times and the roller coaster 22 times, using a total of 2020 tickets. This gives us the first equation:\newline5x+2y=205x + 2y = 20
  2. First Equation: Then, Stanley went on the Ferris wheel 22 times and the roller coaster 55 times, using a total of 2929 tickets. This gives us the second equation:\newline2x+5y=292x + 5y = 29
  3. Second Equation: We now have a system of equations:\newline5x+2y=205x + 2y = 20\newline2x+5y=292x + 5y = 29\newlineWe can solve this system using either substitution or elimination. Let's use the elimination method.
  4. Elimination Method: Multiply the first equation by 22 and the second equation by 55 to make the coefficients of xx the same:\newline(5x+2y)×2=20×2(5x + 2y) \times 2 = 20 \times 2\newline(2x+5y)×5=29×5(2x + 5y) \times 5 = 29 \times 5\newlineThis gives us:\newline10x+4y=4010x + 4y = 40\newline10x+25y=14510x + 25y = 145
  5. Multiply Equations: Subtract the second new equation from the first new equation to eliminate xx:(10x+4y)−(10x+25y)=40−145(10x + 4y) - (10x + 25y) = 40 - 14510x+4y−10x−25y=−10510x + 4y - 10x - 25y = -105−21y=−105-21y = -105
  6. Eliminate xx: Solve for yy:\(\newline\)−21-21yy = −105-105\(\newline\)yy = −105-105 / −21-21\(\newline\)yy = 55
  7. Solve for y: Now that we have the value for yy, we can substitute it back into one of the original equations to solve for xx. Let's use the first equation:\newline5x+2y=205x + 2y = 20\newline5x+2(5)=205x + 2(5) = 20\newline5x+10=205x + 10 = 20
  8. Substitute into Equation: Solve for xx:5x+10=205x + 10 = 205x=20−105x = 20 - 105x=105x = 10x=105x = \frac{10}{5}x=2x = 2

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