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Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineStudents in a health class are tracking how much water they consume each day. Melissa has two reusable water bottles: a small one and a large one. Yesterday, she drank 22 small bottles and 44 large bottles, for a total of 2,7942,794 grams. The day before, she drank 11 small bottle and 44 large bottles, for a total of 2,4112,411 grams. How much does each bottle hold?\newlineThe small bottle holds _____ grams and the large one holds _____ grams.

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineStudents in a health class are tracking how much water they consume each day. Melissa has two reusable water bottles: a small one and a large one. Yesterday, she drank 22 small bottles and 44 large bottles, for a total of 2,7942,794 grams. The day before, she drank 11 small bottle and 44 large bottles, for a total of 2,4112,411 grams. How much does each bottle hold?\newlineThe small bottle holds _____ grams and the large one holds _____ grams.
  1. Set up equations: Let's denote the amount of grams the small bottle holds as SS and the large bottle as LL. We can set up two equations based on the given information:\newline11. For the first day: 2S+4L=27942S + 4L = 2794 grams\newline22. For the second day: 1S+4L=24111S + 4L = 2411 grams
  2. Use elimination method: To use elimination, we can multiply the second equation by 22 to align the coefficients of SS with the first equation:\newline2(1S+4L)=2×24112(1S + 4L) = 2 \times 2411\newlineThis gives us: 2S+8L=48222S + 8L = 4822 grams
  3. Subtract equations: Now we subtract the first equation from this new equation to eliminate SS:(2S+8L)−(2S+4L)=4822−2794(2S + 8L) - (2S + 4L) = 4822 - 2794This simplifies to: 4L=20284L = 2028 grams
  4. Solve for L: We can now solve for L by dividing both sides by 44:\newline4L4=20284\frac{4L}{4} = \frac{2028}{4}\newlineL=507L = 507 grams
  5. Substitute back and solve for S: With the value of L known, we can substitute it back into either of the original equations to solve for S. We'll use the second equation:\newline1S+4(507)=24111S + 4(507) = 2411\newlineS+2028=2411S + 2028 = 2411
  6. Final solution: Subtract 20282028 from both sides to solve for SS: \newlineS=2411−2028S = 2411 - 2028\newlineS=383S = 383 grams

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