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Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineA librarian is expanding some sections of the city library. He buys books at a special price from a dealer who charges one price for any hardback book and another price for any paperback book. For the children's section, Mr. Crosby purchased 2828 new hardcover books and 3838 new paperback books, which cost a total of $506\$506. He also purchased 5858 new hardcover books and 3838 new paperback books for the adult fiction section, spending a total of $926\$926. What is the special price for each type of book?\newlineThe special price is $____\$\_\_\_\_ for hardcover books and $____\$\_\_\_\_ for paperback books.

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineA librarian is expanding some sections of the city library. He buys books at a special price from a dealer who charges one price for any hardback book and another price for any paperback book. For the children's section, Mr. Crosby purchased 2828 new hardcover books and 3838 new paperback books, which cost a total of $506\$506. He also purchased 5858 new hardcover books and 3838 new paperback books for the adult fiction section, spending a total of $926\$926. What is the special price for each type of book?\newlineThe special price is $____\$\_\_\_\_ for hardcover books and $____\$\_\_\_\_ for paperback books.
  1. Define variables: Step 11: Define the variables for the prices of the books.\newlineLet xx be the price of one hardcover book and yy be the price of one paperback book.
  2. Write equations: Step 22: Write the equations based on the information given.\newlineFor the children's section: 2828 hardcover books and 3838 paperback books cost $506\$506.\newlineEquation: 28x+38y=50628x + 38y = 506\newlineFor the adult fiction section: 5858 hardcover books and 3838 paperback books cost $926\$926.\newlineEquation: 58x+38y=92658x + 38y = 926
  3. Use elimination: Step 33: Use elimination to solve the system of equations.\newlineWe can eliminate yy by subtracting the first equation from the second.\newline(58x+38y)−(28x+38y)=926−506(58x + 38y) - (28x + 38y) = 926 - 506\newline30x=42030x = 420
  4. Solve for x: Step 44: Solve for x.\newlineDivide both sides by 3030 to find x.\newlinex=42030x = \frac{420}{30}\newlinex=14x = 14
  5. Substitute and solve for y: Step 55: Substitute x=14x = 14 back into one of the original equations to solve for y.\newlineUsing the first equation: 28(14)+38y=50628(14) + 38y = 506\newline392+38y=506392 + 38y = 506\newline38y=506−39238y = 506 - 392\newline38y=11438y = 114\newliney=11438y = \frac{114}{38}\newliney=3y = 3

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