Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ A coffee shop is having a sale on prepackaged coffee and tea. Yesterday they sold $14$ packages of coffee and $36$ packages of tea, for which customers paid a total of $\$386$ . The day before, $20$ packages of coffee and $36$ packages of tea was sold, which brought in a total of $\$428$ . How much does each package cost? $\newline$ Per package, coffee costs $\$$ _ and tea costs $\$$ _.

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Algebra 1

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

A coffee shop is having a sale on prepackaged coffee and tea. Yesterday they sold

14

packages of coffee and

36

packages of tea, for which customers paid a total of

\$386

. The day before,

20

packages of coffee and

36

packages of tea was sold, which brought in a total of

\$428

. How much does each package cost?

\newline

Per package, coffee costs

\$

_____ and tea costs

\$

_____.

Define Variables: Let's denote the cost of one package of coffee as $x$ dollars and the cost of one package of tea as $y$ dollars. We can write two equations based on the information given for the two days. $\newline$ Yesterday's sale: $14$ packages of coffee and $36$ packages of tea for $\$386$ . $\newline$ The day before's sale: $20$ packages of coffee and $36$ packages of tea for $\$428$ .
Write Equations: Write the equations based on the information given. $\newline$ For yesterday's sale: $14x + 36y = 386$ $\newline$ For the day before's sale: $20x + 36y = 428$
Eliminate Variable: We have the system of equations: $\newline$ $14x + 36y = 386$ $\newline$ $20x + 36y = 428$ $\newline$ We need to eliminate one of the variables. Since the coefficients of $y$ are the same in both equations, we can eliminate $y$ by subtracting the first equation from the second.
Subtract Equations: Subtract the first equation from the second to eliminate $y$ . $\newline$ $(20x + 36y) - (14x + 36y) = 428 - 386$ $\newline$ $20x - 14x + 36y - 36y = 428 - 386$ $\newline$ $6x = 42$
Solve for x: Solve for x. $\newline$ $6x = 42$ $\newline$ $x = \frac{42}{6}$ $\newline$ $x = 7$ $\newline$ So, each package of coffee costs $\$7$ .
Substitute $x$ : Now that we have the value of $x$ , we can substitute it back into one of the original equations to find the value of $y$ . Let's use the first equation: $14x + 36y = 386$ Substitute $x = 7$ into the equation: $14(7) + 36y = 386$ $98 + 36y = 386$
Solve for y: Solve for y. $\newline$ $36y = 386 - 98$ $\newline$ $36y = 288$ $\newline$ $y = \frac{288}{36}$ $\newline$ $y = 8$ $\newline$ So, each package of tea costs $\$8$ .