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Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineAn employee at a company that assembles chandeliers is packing boxes for shipping. In the first box, he packed 22 small chandeliers and 55 large chandeliers, which weighed a total of 318318 pounds. In the second box, he packed 33 small chandeliers and 44 large chandeliers, which had a weight of 295295 pounds. Assuming the weight of the box isn't included in the shipping weight, how much does each size of chandelier weigh?\newlineEach small chandelier weighs _\_ pounds and each large one weighs _\_ pounds.

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineAn employee at a company that assembles chandeliers is packing boxes for shipping. In the first box, he packed 22 small chandeliers and 55 large chandeliers, which weighed a total of 318318 pounds. In the second box, he packed 33 small chandeliers and 44 large chandeliers, which had a weight of 295295 pounds. Assuming the weight of the box isn't included in the shipping weight, how much does each size of chandelier weigh?\newlineEach small chandelier weighs _\_ pounds and each large one weighs _\_ pounds.
  1. Define variables: Define the variables for the weights of the chandeliers.\newlineLet xx be the weight of a small chandelier and yy be the weight of a large chandelier.
  2. Write equations: Write the equations based on the given information.\newlineFor the first box: 22 small chandeliers and 55 large chandeliers weigh 318318 pounds.\newline2x+5y=3182x + 5y = 318\newlineFor the second box: 33 small chandeliers and 44 large chandeliers weigh 295295 pounds.\newline3x+4y=2953x + 4y = 295
  3. Eliminate variable: Choose which variable to eliminate.\newlineWe will eliminate xx by multiplying the first equation by −3-3 and the second equation by 22 to make the coefficients of xx opposites.
  4. Multiply equations: Multiply the equations.\newlineFirst equation multiplied by −3-3:\newline−3(2x+5y)=−3(318)-3(2x + 5y) = -3(318)\newline−6x−15y=−954-6x - 15y = -954\newlineSecond equation multiplied by 22:\newline2(3x+4y)=2(295)2(3x + 4y) = 2(295)\newline6x+8y=5906x + 8y = 590
  5. Add equations: Add the new equations to eliminate xx.
    (−6x−15y)+(6x+8y)=−954+590(-6x - 15y) + (6x + 8y) = -954 + 590
    −6x+6x−15y+8y=−954+590-6x + 6x - 15y + 8y = -954 + 590
    −7y=−364-7y = -364
  6. Solve for y: Solve for y.\newline−7y=−364-7y = -364\newlineDivide both sides by −7-7:\newliney=−364−7y = \frac{-364}{-7}\newliney=52y = 52
  7. Substitute and solve: Substitute yy back into one of the original equations to solve for xx. Using the first equation 2x+5y=3182x + 5y = 318: 2x+5(52)=3182x + 5(52) = 318 2x+260=3182x + 260 = 318
  8. Solve for x: Solve for x.\newlineSubtract 260260 from both sides:\newline2x+260−260=318−2602x + 260 - 260 = 318 - 260\newline2x=582x = 58\newlineDivide both sides by 22:\newlinex=582x = \frac{58}{2}\newlinex=29x = 29

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