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Gnoma [55]
4 years ago
13

At a candy store, Jenny bought 1 pound of jelly beans and 5 pounds of gummy worms for $41. Meanwhile, Ozzie bought 6 pounds of j

elly beans and 2 pounds of gummy worms for $50. How much does the candy cost?
Mathematics
2 answers:
natka813 [3]4 years ago
8 0
91 dollars? I did not understand the question
Eva8 [605]4 years ago
8 0
_Award brainliest if i helped thx! 

Let Jelly Beans be X and Gummy Worms be Y
X + 5 Y = 41   , X = 41-5Y
6X + 2 Y = 50 
Substitute X =41-5Y 
6(41-5Y) + 2Y = 50
Y = 7 Hence X = 6
1 Pound of : Jelly Beans Cost $6, Gummy Worms cost $7
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Question 5 (5 points)
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The solution of the system of equations is, (1,-1,2)

Step-by-step explanation:

Given system equation;

x + 5y - 3z = -10

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\left[\begin{array}{ccc}1&5&-3\\-5&6&-5\\-1&8&-8\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}-10\\-21\\-25\end{array}\right] \\\\\\det. = 1\left[\begin{array}{cc}\\6&-5\\8&-8\end{array}\right] -5\left[\begin{array}{cc}\\-5&-5\\-1&-8\end{array}\right] -3\left[\begin{array}{cc}\\-5&6\\-1&8\end{array}\right] \\\\\\det. = 1(-8) -5(35)-3(-34)= -8 - 175+ 102 = -81

Cofactor;

First \ row \left[\begin{array}{cc}+\\ 6&-5\\\ 8&-8\end{array}\right  \left\begin{array}{cc}-\\ -5&-5\\-1&-8\end{array}\right \left\begin{array}{cc}+\\-5&6\\-1&8\end{array}\right] = [-8  \ \ -35 \ \ -34]\\\\\\\ Second \ row \left[\begin{array}{cc}-\\ 5&-3\\\ 8&-8\end{array}\right  \left\begin{array}{cc}+\\ 1&-3\\-1&-8\end{array}\right \left\begin{array}{cc}-\\1&5\\-1&8\end{array}\right]  = [16\ \ -11 \ \ -13]\\\\\\

Third \ row \left[\begin{array}{cc}+\\ 5&-3\\\ 6&-5\end{array}\right  \left\begin{array}{cc}-\\ 1&-3\\-5&-5\end{array}\right \left\begin{array}{cc}+\\1&5\\-5&6\end{array}\right]= [-7 \  \ 20\ \ 31]

Cofactor = \left[\begin{array}{ccc}-8&-35&-34\\16&-11&-13\\-7&20&31\end{array}\right]

inverse \ matrix =-\frac{1}{81}  \left[\begin{array}{ccc}-8&16&-7\\-35&-11&20\\-34&-13&31\end{array}\right] \\\\\\

Solution of the matrix:

\left[\begin{array}{c}x\\y\\z\end{array}\right] = -\frac{1}{81}  \left[\begin{array}{ccc}-8&16&-7\\-35&-11&20\\-34&-13&31\end{array}\right]  X \left[\begin{array}{c}-10\\-21\\-25\end{array}\right] = \left[\begin{array}{c}\frac{-8*-10 }{-81 } +\frac{16*-21 }{-81 } + \frac{-7*-25 }{-81 }\\\\\frac{-35*-10 }{-81 } +\frac{-11*-21 }{-81 }+ \frac{20*-25 }{-81 }\\\\\frac{-34*-10 }{-81 }+ \frac{-13*-21 }{-81 }+ \frac{31*-25 }{-81 }\end{array}\right] \\\\\

\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}\frac{-81}{-81} \\\\\frac{81}{-81} \\\\\frac{-162}{-81} \end{array}\right] =  \left[\begin{array}{c}1\\-1\\2\end{array}\right]

Therefore, the correct option is (1,-1,2)

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