Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ Band students at Silvergrove High School sell candy every year as a fundraiser. Last year, they sold $86$ boxes of truffles and $87$ boxes of peanut brittle, raising a total of $\$519$ . This year, they sold $62$ boxes of truffles and $74$ boxes of peanut brittle, from which they raised $\$408$ . How much does the band earn from each item? $\newline$ The band earns $\$\_$ from each box of truffles and $\$\_$ from each box of peanut brittle.

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Grade 8

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

Band students at Silvergrove High School sell candy every year as a fundraiser. Last year, they sold

86

boxes of truffles and

87

boxes of peanut brittle, raising a total of

\$519

. This year, they sold

62

boxes of truffles and

74

boxes of peanut brittle, from which they raised

\$408

. How much does the band earn from each item?

\newline

The band earns

\$\_

from each box of truffles and

\$\_

from each box of peanut brittle.

Define variables: Let's define two variables: let $x$ be the amount the band earns from each box of truffles, and $y$ be the amount the band earns from each box of peanut brittle. We can then write two equations based on the information given: $\newline$ $1$ . For last year's sales: $86x + 87y = 519$ $\newline$ $2$ . For this year's sales: $62x + 74y = 408$ $\newline$ These two equations form our system of equations.
Eliminate variable y: To solve this system using elimination, we need to eliminate one of the variables. We can do this by multiplying the equations by numbers that will make the coefficients of one of the variables the same. Let's try to eliminate $y$ by finding a common multiple of $87$ and $74$ . The least common multiple of $87$ and $74$ is $87 \times 74$ , but we can use smaller multipliers to keep the numbers manageable. We can multiply the first equation by $74$ and the second equation by $87$ to get the coefficients of $y$ to match: $\newline$ $74(86x + 87y) = 74 \times 519$ $\newline$ $87$ $0$ $\newline$ This gives us: $\newline$ $87$ $1$ $\newline$ $87$ $2$
Perform multiplication: Now let's perform the multiplication: $\newline$ $6364x + 6438y = 38406$ $\newline$ $5394x + 6438y = 35496$ $\newline$ We now have two new equations: $\newline$ $1$ . $6364x + 6438y = 38406$ $\newline$ $2$ . $5394x + 6438y = 35496$
Subtract equations: Next, we subtract the second equation from the first to eliminate $y$ : $(6364x + 6438y) - (5394x + 6438y) = 38406 - 35496$ This simplifies to: $970x = 2910$
Solve for x: Now we divide both sides by $970$ to solve for $x$ : $\frac{970x}{970} = \frac{2910}{970}$ $x = 3$
Substitute $x$ into equation: Now that we have the value for $x$ , we can substitute it back into one of the original equations to solve for $y$ . Let's use the first equation: $\newline$ $86x + 87y = 519$ $\newline$ Substitute $x = 3$ : $\newline$ $86(3) + 87y = 519$ $\newline$ $258 + 87y = 519$
Solve for y: Subtract $258$ from both sides to solve for y: $\newline$ $87y = 519 - 258$ $\newline$ $87y = 261$
Final solution: Now we divide both sides by $87$ to solve for $y$ : $\newline$ $\frac{87y}{87} = \frac{261}{87}$ $\newline$ $y = 3$
Final solution: Now we divide both sides by $87$ to solve for $y$ : $\newline$ $\frac{87y}{87} = \frac{261}{87}$ $\newline$ $y = 3$ We have found that $x = 3$ and $y = 3$ , which means the band earns $\$3$ from each box of truffles and $\$3$ from each box of peanut brittle.