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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. Javier has two reusable water bottles: a small one and a large one. Yesterday, he drank 22 small bottles and 11 large bottle, for a total of 1,2301,230 grams. The day before, he drank 44 small bottles and 33 large bottles, for a total of 2,9382,938 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. Javier has two reusable water bottles: a small one and a large one. Yesterday, he drank 22 small bottles and 11 large bottle, for a total of 1,2301,230 grams. The day before, he drank 44 small bottles and 33 large bottles, for a total of 2,9382,938 grams. How much does each bottle hold?\newlineThe small bottle holds _\_ grams and the large one holds _\_ grams.
  1. Define Equations: Let's denote the amount of grams the small bottle holds as SS and the large bottle as LL. We can then write two equations based on the given information.
  2. First Day Consumption: The first equation comes from the first day's consumption: 2S+1L=1,2302S + 1L = 1,230 grams.
  3. Second Day Consumption: The second equation comes from the second day's consumption: 4S+3L=2,9384S + 3L = 2,938 grams.
  4. Elimination Method: To use elimination, we need to manipulate the equations so that when we add or subtract them, one of the variables is eliminated. Let's multiply the first equation by 33 to align the LL terms.\newline3(2S+1L)=3(1,230)3(2S + 1L) = 3(1,230)\newlineThis gives us: 6S+3L=3,6906S + 3L = 3,690
  5. New Equations: Now we have a system of two new equations:\newline6S+3L=3,6906S + 3L = 3,690\newline4S+3L=2,9384S + 3L = 2,938
  6. Eliminate Variable: Subtract the second equation from the first to eliminate LL:(6S+3L)−(4S+3L)=3,690−2,938(6S + 3L) - (4S + 3L) = 3,690 - 2,938This simplifies to: 2S=7522S = 752
  7. Solve for S: Now, divide both sides by 22 to solve for S:\newline2S2=7522\frac{2S}{2} = \frac{752}{2}\newlineS=376S = 376
  8. Substitute for L: Now that we have the value for SS, we can substitute it back into one of the original equations to solve for LL. Let's use the first equation:\newline2(376)+L=1,2302(376) + L = 1,230\newline752+L=1,230752 + L = 1,230
  9. Final Solution: Subtract 752752 from both sides to solve for LL: \newlineL=1,230−752L = 1,230 - 752\newlineL=478L = 478

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