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Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineLivingston's Bakery sold one customer 88 dozen chocolate cookies and 1010 dozen oatmeal cookies for $104\$104. The bakery also sold another customer 44 dozen chocolate cookies and 33 dozen oatmeal cookies for $44\$44. How much do the cookies cost?\newlineA dozen chocolate cookies cost $____\$\_\_\_\_, and a dozen oatmeal cookies cost $____\$\_\_\_\_.

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Q. Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.\newlineLivingston's Bakery sold one customer 88 dozen chocolate cookies and 1010 dozen oatmeal cookies for $104\$104. The bakery also sold another customer 44 dozen chocolate cookies and 33 dozen oatmeal cookies for $44\$44. How much do the cookies cost?\newlineA dozen chocolate cookies cost $____\$\_\_\_\_, and a dozen oatmeal cookies cost $____\$\_\_\_\_.
  1. Define Cookie Costs: Let's denote the cost of a dozen chocolate cookies as c c dollars and the cost of a dozen oatmeal cookies as o o dollars.\newlineThe first customer bought 88 dozen chocolate cookies and 1010 dozen oatmeal cookies for \(104\). This can be represented by the equation:\(\newline\)\( 8c + 10o = 104 \)
  2. First Customer Purchase: The second customer bought \(4\) dozen chocolate cookies and \(3\) dozen oatmeal cookies for 4444. This can be represented by the equation:\newline4c+3o=44 4c + 3o = 44
  3. Second Customer Purchase: We now have a system of equations to solve:\newline8c+10o=104 8c + 10o = 104 \newline4c+3o=44 4c + 3o = 44 \newlineWe can use either substitution or elimination to solve this system. Let's use the elimination method.
  4. Solve System of Equations: To eliminate one of the variables, we can multiply the second equation by 22 to make the coefficient of c c the same in both equations:\newline2(4c+3o)=2(44) 2(4c + 3o) = 2(44) \newline8c+6o=88 8c + 6o = 88
  5. Eliminate Variable c: Now we subtract the new equation from the first equation to eliminate c c :\newline(8c+10o)−(8c+6o)=104−88 (8c + 10o) - (8c + 6o) = 104 - 88 \newline4o=16 4o = 16
  6. Find Cost of Oatmeal Cookies: Solving for o o , we get:\newlineo=164 o = \frac{16}{4} \newlineo=4 o = 4 \newlineSo, a dozen oatmeal cookies cost \(4\).
  7. Substitute to Find Cost of Chocolate Cookies: Now we can substitute the value of \( o \) into one of the original equations to find \( c \). Let's use the second equation:\(\newline\)\( 4c + 3(4) = 44 \)\(\newline\)\( 4c + 12 = 44 \)\(\newline\)\( 4c = 44 - 12 \)\(\newline\)\( 4c = 32 \)
  8. Substitute to Find Cost of Chocolate Cookies: Now we can substitute the value of \( o \) into one of the original equations to find \( c \). Let's use the second equation:\(\newline\)\( 4c + 3(4) = 44 \)\(\newline\)\( 4c + 12 = 44 \)\(\newline\)\( 4c = 44 - 12 \)\(\newline\)\( 4c = 32 \)Solving for \( c \), we get:\(\newline\)\( c = \frac{32}{4} \)\(\newline\)\( c = 8 \)\(\newline\)So, a dozen chocolate cookies cost 88.

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