Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks. For a project in statistics class, a pair of students decided to invest in two companies, one that produces software and one that does biotechnology research. Akira purchased $1$ share in the software company, which cost a total of $\$38$ . At the same time, Daniel invested a total of $\$3,312$ in $33$ shares in the software company and $21$ shares in the biotech firm. How much did each share cost? Each share in the software company cost $\$$ $\_\_\_\_$ , and each share in the biotech firm cost $\$$ $\_\_\_\_$ .

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Solve a system of equations using substitution: word problems

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Q. Write a system of equations to describe the situation below, solve using an augmented matrix, and fill in the blanks. For a project in statistics class, a pair of students decided to invest in two companies, one that produces software and one that does biotechnology research. Akira purchased

1

share in the software company, which cost a total of

\$38

. At the same time, Daniel invested a total of

\$3,312

33

shares in the software company and

21

shares in the biotech firm. How much did each share cost? Each share in the software company cost

\$

\_\_\_\_

, and each share in the biotech firm cost

\$

\_\_\_\_

Define Equations: Let's call the cost of a share in the software company $s$ and the cost of a share in the biotech firm $b$ . $\newline$ Akira's purchase can be represented by the equation: $1s = 38$ .
Create Augmented Matrix: Daniel's investment can be represented by the equation: $33s + 21b = 3312$ .
Perform Row Operations: Now we have a system of two equations: $\newline$ $1$ . $1s = 38$ $\newline$ $2$ . $33s + 21b = 3312$ $\newline$ We'll write this system as an augmented matrix.
Calculate New Value: The augmented matrix is: $\newline$ \begin{array}{cc|c} 1 & 0 & 38 \ 33 & 21 & 3312 \ \end{array}
Solve for $b$ : To solve the system using the matrix, we'll perform row operations to get the matrix into reduced row-echelon form. $\newline$ First, we'll multiply the first row by $-33$ and add it to the second row to eliminate the $s$ term from the second equation.
Calculate Share Cost: After the row operation, the new matrix is: $\newline$ $\begin{pmatrix} 1 & 0 & 38 \ 0 & 21 & 3312 - (33 \times 38) \end{pmatrix}$
Final Solution: Now we calculate $3312 - (33 \times 38)$ to find the new value for the second row, second column.
Final Solution: Now we calculate $3312 - (33 \times 38)$ to find the new value for the second row, second column.The calculation gives us $3312 - (33 \times 38) = 3312 - 1254 = 2058$ . So the updated matrix is: $\begin{vmatrix} 1 & 0 & 38 \ 0 & 21 & 2058 \end{vmatrix}$
Final Solution: Now we calculate $3312 - (33 \times 38)$ to find the new value for the second row, second column.The calculation gives us $3312 - (33 \times 38) = 3312 - 1254 = 2058$ . So the updated matrix is: $\begin{vmatrix} 1 & 0 & 38 \ 0 & 21 & 2058 \end{vmatrix}$ Next, we divide the second row by $21$ to solve for $b$ .
Final Solution: Now we calculate $3312 - (33 \times 38)$ to find the new value for the second row, second column.The calculation gives us $3312 - (33 \times 38) = 3312 - 1254 = 2058$ . So the updated matrix is: $\begin{vmatrix} 1 & 0 & 38 \ 0 & 21 & 2058 \end{vmatrix}$ Next, we divide the second row by $21$ to solve for $b$ . Dividing $2058$ by $21$ gives us $b = \frac{2058}{21} = 98$ . So, each share in the biotech firm costs \$(\$$98$)\$.
Final Solution: Now we calculate $3312 - (33 \times 38)$ to find the new value for the second row, second column.The calculation gives us $3312 - (33 \times 38) = 3312 - 1254 = 2058$. So the updated matrix is: $\begin{vmatrix} 1 & 0 & 38 \ 0 & 21 & 2058 \end{vmatrix}$ Next, we divide the second row by $21$ to solve for $b$. Dividing $2058$ by $21$ gives us $b = \frac{2058}{21} = 98$. So, each share in the biotech firm costs \$$98$. Now that we have the value for $b$, we can use the first equation $1s = 38$ to solve for $3312 - (33 \times 38) = 3312 - 1254 = 2058$$0$.
Final Solution: Now we calculate $3312 - (33 \times 38)$ to find the new value for the second row, second column.The calculation gives us $3312 - (33 \times 38) = 3312 - 1254 = 2058$. So the updated matrix is: $\begin{vmatrix} 1 & 0 & 38 \ 0 & 21 & 2058 \end{vmatrix}$ Next, we divide the second row by $21$ to solve for $b$. Dividing $2058$ by $21$ gives us $b = \frac{2058}{21} = 98$. So, each share in the biotech firm costs \$(\$$98$)$. Now that we have the value for $b$, we can use the first equation $$1$s = $38$$3312 - (33 \times 38) = 3312 - 1254 = 2058$$0$s$3312 - (33 \times 38) = 3312 - 1254 = 2058$$1$s = $38$$3312 - (33 \times 38) = 3312 - 1254 = 2058$$2$.