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tiny-mole [99]
3 years ago
12

A local sorority sold hot dogs and bratwursts at the spring fling picnics. The first day they sold 8 dozen hot dogs and 13 dozen

bratwursts for ​$282.60. The second day they sold 10 dozen hot dogs and 15 dozen bratwursts for a total of ​$333.00. How much did each​ cost? Solve this system using an inverse matrix.​ (Hint: It's easier to leave the dozens in the problem until the last​ step.)
Mathematics
1 answer:
kicyunya [14]3 years ago
4 0

Answer:

1 hot dog costs $0.75

1 bratwurst costs $1.35

Step-by-step explanation:

Let x and y be the price per dozen of hot dogs and bratwursts respectively.

The first day they sold 8 dozen hot dogs and 13 dozen bratwursts for ​$282.60

8x + 13y = 282.60

The second day they sold 10 dozen hot dogs and 15 dozen bratwursts for a total of ​$333.00

10x + 15y = 333

and we have the linear system

<em>8x + 13y = 282.60 </em>

<em>10x + 15y = 333 </em>

which can be written in matrix form as

\bf \left(\begin{array}{cc}8&13\\10&15\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}282.60\\333\end{array}\right)

The solution would be given by

\bf \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}8&13\\10&15\end{array}\right)^{-1}\left(\begin{array}{c}282.60\\333\end{array}\right)

We have  

\bf \left(\begin{array}{cc}8&13\\10&15\end{array}\right)^{-1}=\left(\begin{array}{cc}-3/2&13/10\\1&-4/5\end{array}\right)

hence

\bf \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}-3/2&13/10\\1&-4/5\end{array}\right)\left(\begin{array}{c}282.60\\333\end{array}\right)=\left(\begin{array}{c}9\\ 16.2\end{array}\right)

Now,

if a dozen hot dogs cost $9, 1 hot dog costs 9/12 = $0.75

if a dozen bratwursts cost $16.2, 1 bratwurst costs 16.2/12 = $1.35

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