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 $4$ small chandeliers and $2$ large chandeliers, which weighed a total of $66$ kilograms. In the second box, he packed $4$ small chandeliers and $4$ large chandeliers, which had a weight of $112$ kilograms. 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 $\_$ kilograms and each large one weighs $\_$ kilograms.

Full solution

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

4

small chandeliers and

2

large chandeliers, which weighed a total of

66

kilograms. In the second box, he packed

4

small chandeliers and

4

large chandeliers, which had a weight of

112

kilograms. 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

\_

kilograms and each large one weighs

\_

kilograms.

Denote weights: Let's denote the weight of each small chandelier as $s$ and the weight of each large chandelier as $l$ . The first box with $4$ small chandeliers and $2$ large chandeliers weighs a total of $66$ kilograms, which gives us the equation $4s + 2l = 66$ .
First box equation: The second box with $4$ small chandeliers and $4$ large chandeliers weighs $112$ kilograms, which gives us the equation $4s + 4l = 112$ .
Second box equation: We now have a system of two equations. To use elimination, we can subtract the first equation from the second equation to eliminate $s$ . Subtracting $4s + 2l = 66$ from $4s + 4l = 112$ gives us $2l = 46$ .
Elimination method: Dividing both sides of $2l = 46$ by $2$ gives us $l = 23$ . This means each large chandelier weighs $23$ kilograms.
Large chandelier weight: Now that we know the weight of each large chandelier, we can substitute $l = 23$ into the first equation $4s + 2l = 66$ to find the weight of each small chandelier. Substituting $l$ gives us $4s + 2(23) = 66$ .
Substitute in first equation: Simplifying the equation $4s + 46 = 66$ by subtracting $46$ from both sides gives us $4s = 20$ .
Find small chandelier weight: Dividing both sides of $4s = 20$ by $4$ gives us $s = 5$ . This means each small chandelier weighs $5$ kilograms.