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 $48$ small gifts and $29$ large gifts, earning a total of $\$318$ . Today, they wrapped $19$ small gifts and $43$ large gifts, and earned $\$315$ . 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

48

small gifts and

29

large gifts, earning a total of

\$318

. Today, they wrapped

19

small gifts and

43

large gifts, and earned

\$315

. 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 variables: Define the variables for the cost to wrap a small gift and a large gift. Let $x$ be the cost to wrap a small gift, and $y$ be the cost to wrap a large gift.
Write first day's earnings equation: Write the equation for the first day's earnings. $\newline$ $48$ small gifts and $29$ large gifts earned a total of $\$318$ . $\newline$ $48x + 29y = 318$
Write second day's earnings equation: Write the equation for the second day's earnings. $\newline$ $19$ small gifts and $43$ large gifts earned a total of $\$315$ . $\newline$ $19x + 43y = 315$
Decide variable to eliminate: Decide which variable to eliminate. $\newline$ We can choose to eliminate either $x$ or $y$ . To make calculations simpler, we will eliminate $x$ by multiplying the first equation by $19$ and the second equation by $48$ , because $19$ and $48$ are the coefficients of $x$ in the two equations.
Multiply equations by coefficients: Multiply the first equation by $19$ and the second equation by $48$ . $\newline$ First equation: $(48x + 29y) \times 19 = 318 \times 19$ $\newline$ Second equation: $(19x + 43y) \times 48 = 315 \times 48$
Write new equations after multiplication: Write the new equations after multiplication. $\newline$ First equation: $912x + 551y = 6042$ $\newline$ Second equation: $912x + 2064y = 15120$
Subtract equations to eliminate x: Subtract the second equation from the first to eliminate x. $\newline$ $(912x + 551y) - (912x + 2064y) = 6042 - 15120$
Perform subtraction to solve for y: Perform the subtraction to solve for y. $\newline$ $912x + 551y - 912x - 2064y = 6042 - 15120$ $\newline$ $551y - 2064y = 6042 - 15120$ $\newline$ $-1513y = -9078$
Solve for y: Solve for y. $\newline$ $y = \frac{-9078}{-1513}$ $\newline$ $y = 6$
Substitute $y$ into original equation: Substitute $y = 6$ into one of the original equations to solve for $x$ . Using the first day's equation: $48x + 29(6) = 318$
Perform substitution and solve for $x$ : Perform the substitution and solve for $x$ . $48x + 174 = 318$ $48x = 318 - 174$ $48x = 144$
Divide to find $x$ : Divide both sides by $48$ to find $x$ . $x = \frac{144}{48}$ $x = 3$
Final solution: We found $x = 3$ and $y = 6$ . The organization charges $\$3$ to wrap a small gift and $\$6$ to wrap a large one.