Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ At a historical landmark, candles are used to simulate an authentic atmosphere. A volunteer is currently putting new candles in the candle holders. On the east side, he replaced candles in $7$ small candle holders and $3$ large candle holders, using a total of $39$ candles. On the west side, he replaced the candles in $23$ small candle holders and $7$ large candle holders, for a total of $111$ candles. How many candles does each candle holder hold? $\newline$ Each small candleholder holds $\_$ candles, and each large one holds $\_$ candles.

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Algebra 2

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

At a historical landmark, candles are used to simulate an authentic atmosphere. A volunteer is currently putting new candles in the candle holders. On the east side, he replaced candles in

7

small candle holders and

3

large candle holders, using a total of

39

candles. On the west side, he replaced the candles in

23

small candle holders and

7

large candle holders, for a total of

111

candles. How many candles does each candle holder hold?

\newline

Each small candleholder holds

\_

candles, and each large one holds

\_

candles.

Define variables: Define variables for the number of candles each type of candle holder can hold. Let $s$ be the number of candles a small candle holder holds, and $l$ be the number of candles a large candle holder holds. The first equation from the east side is $7s + 3l = 39$ .
Create second equation: Create the second equation from the west side data, which is $23s + 7l = 111$ .
Eliminate variable: To eliminate one variable using the elimination method, we aim to eliminate $l$ . Multiply the first equation by $7$ (the coefficient of $l$ in the second equation) to get $49s + 21l = 273$ .
Multiply equations: Multiply the second equation by $3$ (the coefficient of $l$ in the first equation) to get $69s + 21l = 333$ .
Subtract equations: Subtract the new first equation from the new second equation to eliminate $l$ : $(69s + 21l) - (49s + 21l) = 333 - 273$ , which simplifies to $20s = 60$ .
Solve for s: Solve for s by dividing both sides by $20$ : $s = \frac{60}{20} = 3$ .
Substitute $s$ : Substitute $s = 3$ back into the first original equation to solve for $l$ : $7(3) + 3l = 39$ , which simplifies to $21 + 3l = 39$ .
Solve for l: Solve for $l$ : $3l = 39 - 21$ , $3l = 18$ , then $l = 18 / 3 = 6$ .