Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ An employee at a company that assembles chandeliers is packing boxes for shipping. In the first box, he packed $1$ small chandelier and $4$ large chandeliers, which weighed a total of $207$ pounds. In the second box, he packed $4$ small chandeliers and $1$ large chandelier, which had a weight of $93$ pounds. Assuming the weight of the box isn't included in the shipping weight, how much does each size of chandelier weigh? $\newline$ Each small chandelier weighs $\_$ pounds and each large one weighs $\_$ pounds.

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Solve a system of equations using elimination: word problems

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.

\newline

An employee at a company that assembles chandeliers is packing boxes for shipping. In the first box, he packed

1

small chandelier and

4

large chandeliers, which weighed a total of

207

pounds. In the second box, he packed

4

small chandeliers and

1

large chandelier, which had a weight of

93

pounds. Assuming the weight of the box isn't included in the shipping weight, how much does each size of chandelier weigh?

\newline

Each small chandelier weighs

\_

pounds and each large one weighs

\_

pounds.

Equation $1$ : Let's denote the weight of each small chandelier as $s$ and the weight of each large chandelier as $l$ . The first box with $1$ small chandelier and $4$ large chandeliers weighs a total of $207$ pounds, which gives us the equation $s + 4l = 207$ .
Equation $2$ : The second box with $4$ small chandeliers and $1$ large chandelier weighs $93$ pounds, which gives us the equation $4s + l = 93$ .
Elimination Choice: We now have a system of two equations. To use elimination, we need to eliminate one of the variables, $s$ or $l$ . We choose to eliminate $l$ because its coefficients are multiples of each other.
New Second Equation: To eliminate $l$ , we multiply the second equation by $4$ , the coefficient of $l$ in the first equation. This gives us the new equation $16s + 4l = 372$ .
Elimination Step: We now subtract the first equation from the new second equation to eliminate $l$ . This gives us $16s + 4l - (s + 4l) = 372 - 207$ , which simplifies to $15s = 165$ .
Calculate Small Chandelier Weight: Dividing both sides of the equation by $15$ gives us $s = \frac{165}{15}$ , which simplifies to $s = 11$ . So each small chandelier weighs $11$ pounds.
Substitute and Solve: We substitute $s = 11$ into the first equation and solve for $l$ . This gives us $11 + 4l = 207$ . Subtracting $11$ from both sides gives us $4l = 196$ .
Calculate Large Chandelier Weight: Dividing both sides of the equation by $4$ gives us $l = 196 / 4$ , which simplifies to $l = 49$ . So each large chandelier weighs $49$ pounds.