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 $34$ small gifts and $48$ large gifts, earning a total of $\$486$ . Today, they wrapped $28$ small gifts and $50$ large gifts, and earned $\$484$ . 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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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

34

small gifts and

48

large gifts, earning a total of

\$486

. Today, they wrapped

28

small gifts and

50

large gifts, and earned

\$484

. 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: Let's define the variables for the amount charged to wrap a small gift and a large gift. Let $x$ be the amount charged for wrapping a small gift, and $y$ be the amount charged for wrapping a large gift.
Write equations for days: Now we can write two equations based on the information given for the two days. For yesterday, the equation is $34x + 48y = 486$ . For today, the equation is $28x + 50y = 484$ .
Use elimination method: We will use the elimination method to solve this system of equations. To eliminate one of the variables, we can multiply the first equation by $28$ and the second equation by $34$ , so that the coefficients of $x$ in both equations are the same.
Multiply first equation: Multiplying the first equation $34x + 48y = 486$ by $28$ , we get: $\newline$ $28 \times 34x + 28 \times 48y = 28 \times 486$ $\newline$ $952x + 1344y = 13608$
Multiply second equation: Multiplying the second equation $28x + 50y = 484$ by $34$ , we get: $\newline$ $34 \times 28x + 34 \times 50y = 34 \times 484$ $\newline$ $952x + 1700y = 16456$
Subtract equations: Now we subtract the second equation from the first to eliminate $x$ : $(952x + 1344y) - (952x + 1700y) = 13608 - 16456$ $0x + (1344y - 1700y) = -2848$ $-356y = -2848$
Solve for y: Divide both sides by $-356$ to solve for y: $\newline$ $y = \frac{-2848}{-356}$ $\newline$ $y = 8$
Substitute back to solve $x$ : Now that we have the value for $y$ , we can substitute it back into one of the original equations to solve for $x$ . We'll use the first equation: $34x + 48y = 486$ .
$34x + 48 \times 8 = 486$
$34x + 384 = 486$
Subtract to solve $x$ : Subtract $384$ from both sides to solve for $x$ :
$34x = 486 - 384$
$34x = 102$
Divide to solve $x$ : Divide both sides by $34$ to solve for $x$ :
$x = \frac{102}{34}$
$x = 3$
Final solution: We have found that the organization charges $\$3$ to wrap a small gift and $\$8$ to wrap a large one.