High School

We appreciate your visit to Edna has a home based business making corsages and boutonnieres for school dances Last year she sold 18 corsages and 21 boutonnieres which brought in. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

Edna has a home-based business making corsages and boutonnieres for school dances.

- Last year, she sold 18 corsages and 21 boutonnieres, which brought in a total of $960.
- This year, she sold 25 corsages and 17 boutonnieres, for a total of $1,017.

How much does each item sell for?

Answer :

The answer is:

[tex]\[ \boxed{C = 23, B = 26} \][/tex]

To determine the price of each corsage and boutonniere, we can set up a system of equations based on the information given.

Let [tex]\( C \)[/tex] represent the price of one corsage, and [tex]\( B \)[/tex] represent the price of one boutonniere.

From the first year's sales, we have the equation:

[tex]\[ 18C + 21B = 960 \][/tex] (Equation 1)

From the second year's sales, we have the equation:

[tex]\[ 25C + 17B = 1017 \][/tex] (Equation 2)

Now we have a system of two equations with two unknowns, which we can solve simultaneously.

To solve this system, we can use the method of substitution or elimination. Let's use the elimination method.

First, we can multiply Equation 1 by 25 and Equation 2 by 18 to make the coefficients of [tex]\( C \)[/tex] the same:

[tex]\[ 25 \times (18C + 21B) = 25 \times 960 \][/tex]

[tex]\[ 18 \times (25C + 17B) = 18 \times 1017 \][/tex]

This gives us:

[tex]\[ 450C + 525B = 24000 \][/tex] (Equation 3)

[tex]\[ 450C + 305B = 18306 \][/tex] (Equation 4)

Now, subtract Equation 4 from Equation 3 to eliminate [tex]\( C \)[/tex] :

[tex]\[ (450C + 525B) - (450C + 305B) = 24000 - 18306 \][/tex]

[tex]\[ 220B = 5694 \][/tex]

Divide both sides by 220 to find the price of one boutonniere:

[tex]\[ B = \frac{5694}{220} \][/tex]

[tex]\[ B = 25.88181818 \][/tex]

However, since the prices are likely to be whole numbers, we can round [tex]\( B \)[/tex] to the nearest dollar:

[tex]\[ B \approx 26 \][/tex]

Now, we can substitute [tex]\( B \)[/tex] back into either Equation 1 or Equation 2 to find [tex]\( C \)[/tex]. Let's use Equation 1:

[tex]\[ 18C + 21(26) = 960 \][/tex]

[tex]\[ 18C + 546 = 960 \][/tex]

[tex]\[ 18C = 960 - 546 \][/tex]

[tex]\[ 18C = 414 \][/tex]

Divide both sides by 18 to find the price of one corsage:

[tex]\[ C = \frac{414}{18} \][/tex]

[tex]\[ C = 23 \][/tex]

So, each corsage sells for $23, and each boutonniere sells for $26.

The final answer is:

[tex]\[ \boxed{C = 23, B = 26} \][/tex]

Thanks for taking the time to read Edna has a home based business making corsages and boutonnieres for school dances Last year she sold 18 corsages and 21 boutonnieres which brought in. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada

Answer: Selling price of each corsage = $ 23

Selling price of each boutonniere = $26

Step-by-step explanation:

Let x =Selling price of each corsage.

y = Selling price of each boutonniere.

As per given, we have

[tex]18x+21y= 960\ \ \ \ (i)\\\\ 25x+17y= 1017\ \ \ \ (ii)[/tex]

Divide (i) by 3, we get

[tex]6x+7y=320\\\\\Rightarrow\ 6x=320-7y\\\\\Rightarrow\ x=\dfrac{320-7y}{6}\ \ \ \ (iii)[/tex]

Put this in (ii) , we get

[tex]25(\dfrac{320-7y}{6})+17y=1017\\\\\Rightarrow\ \dfrac{25(320)-25(7y)}{6}+\dfrac{17\times6y}{6}=1017\\\\\Rightarrow\ \dfrac{8000-175y}{6}+\dfrac{102y}{6}=1017\\\\\Rightarrow\ \dfrac{8000-175y+102y}{6}=1017\\\\\Rightarrow\ 8000-73y=6\times1017\\\\\Rightarrow\ 8000-73y=6102\\\\\Rightarrow\ 73y=8000-6102\\\\\Rightarrow\ 73y= 1898\\\\\Rightarrow\ y=\dfrac{1898}{73}\\\\\Rightarrow\ y=26[/tex]

Put this in (iii),

[tex]x=\dfrac{320-7(26)}{6}=23[/tex]

Hence, Selling price of each corsage = $ 23

Selling price of each boutonniere = $26