Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ A local service organization is wrapping gifts at the mall to raise money for charity. Yesterday, they wrapped $15$ small gifts and $33$ large gifts, earning a total of $\$357$ . Today, they wrapped $35$ small gifts and $33$ large gifts, and earned $\$437$ . How much did they charge to wrap the gifts? $\newline$ The organization charges $\_\_\_\_\_$ to wrap a small gift and $\_\_\_\_\_$ to wrap a large one.

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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 local service organization is wrapping gifts at the mall to raise money for charity. Yesterday, they wrapped

15

small gifts and

33

large gifts, earning a total of

\$357

. Today, they wrapped

35

small gifts and

33

large gifts, and earned

\$437

. How much did they charge to wrap the gifts?

\newline

The organization charges

\_\_\_\_\_

to wrap a small gift and

\_\_\_\_\_

to wrap a large one.

Define Costs: Let's denote the cost to wrap a small gift as $x$ dollars and the cost to wrap a large gift as $y$ dollars. We can write two equations based on the information given for the two days. $\newline$ Yesterday's earnings: $15$ small gifts and $33$ large gifts for $\$357$ . $\newline$ Today's earnings: $35$ small gifts and $33$ large gifts for $\$437$ .
Write Equations: Translate the information into two equations. $\newline$ For yesterday: $15x + 33y = 357$ $\newline$ For today: $35x + 33y = 437$
Eliminate Variable: To use elimination, we need to eliminate one of the variables. We can eliminate $y$ by subtracting the first equation from the second equation because they both have the same coefficient for $y$ .
Subtract Equations: Subtract the first equation from the second equation to eliminate $y$ . $\newline$ $(35x + 33y) - (15x + 33y) = 437 - 357$ $\newline$ $35x + 33y - 15x - 33y = 437 - 357$ $\newline$ $20x = 80$
Solve for x: Solve for x by dividing both sides of the equation by $20$ . $\newline$ $\frac{20x}{20} = \frac{80}{20}$ $\newline$ $x = 4$
Substitute $x$ : 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: $15x + 33y = 357$ . $\newline$ $15(4) + 33y = 357$ $\newline$ $60 + 33y = 357$
Solve for y: Subtract $60$ from both sides of the equation to solve for y. $\newline$ $33y = 357 - 60$ $\newline$ $33y = 297$
Solve for y: Subtract $60$ from both sides of the equation to solve for y. $\newline$ $33y = 357 - 60$ $\newline$ $33y = 297$ Divide both sides of the equation by $33$ to find the value of y. $\newline$ $\frac{33y}{33} = \frac{297}{33}$ $\newline$ $y = 9$