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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 11 share in the software company, which cost a total of $38\$38. At the same time, Daniel invested a total of $3,312\$3,312 in 3333 shares in the software company and 2121 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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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 11 share in the software company, which cost a total of $38\$38. At the same time, Daniel invested a total of $3,312\$3,312 in 3333 shares in the software company and 2121 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 $\$____\_\_\_\_.
  1. Define Equations: Let's call the cost of a share in the software company ss and the cost of a share in the biotech firm bb.\newlineAkira's purchase can be represented by the equation: 1s=381s = 38.
  2. Create Augmented Matrix: Daniel's investment can be represented by the equation: 33s+21b=331233s + 21b = 3312.
  3. Perform Row Operations: Now we have a system of two equations:\newline11. 1s=381s = 38\newline22. 33s+21b=331233s + 21b = 3312\newlineWe'll write this system as an augmented matrix.
  4. Calculate New Value: The augmented matrix is:\newline\begin{array}{cc|c} 1 & 0 & 38 \ 33 & 21 & 3312 \ \end{array}
  5. Solve for bb: To solve the system using the matrix, we'll perform row operations to get the matrix into reduced row-echelon form.\newlineFirst, we'll multiply the first row by −33-33 and add it to the second row to eliminate the ss term from the second equation.
  6. Calculate Share Cost: After the row operation, the new matrix is:\newline(1amp;0amp;38 0amp;21amp;3312−(33×38))\begin{pmatrix} 1 & 0 & 38 \ 0 & 21 & 3312 - (33 \times 38) \end{pmatrix}
  7. Final Solution: Now we calculate 3312−(33×38)3312 - (33 \times 38) to find the new value for the second row, second column.
  8. Final Solution: Now we calculate 3312−(33×38)3312 - (33 \times 38) to find the new value for the second row, second column.The calculation gives us 3312−(33×38)=3312−1254=20583312 - (33 \times 38) = 3312 - 1254 = 2058. So the updated matrix is: ∣1amp;0amp;38 0amp;21amp;2058∣\begin{vmatrix} 1 & 0 & 38 \ 0 & 21 & 2058 \end{vmatrix}
  9. Final Solution: Now we calculate 3312−(33×38)3312 - (33 \times 38) to find the new value for the second row, second column.The calculation gives us 3312−(33×38)=3312−1254=20583312 - (33 \times 38) = 3312 - 1254 = 2058. So the updated matrix is: ∣1amp;0amp;38 0amp;21amp;2058∣\begin{vmatrix} 1 & 0 & 38 \ 0 & 21 & 2058 \end{vmatrix} Next, we divide the second row by 2121 to solve for bb.
  10. Final Solution: Now we calculate 3312−(33×38)3312 - (33 \times 38) to find the new value for the second row, second column.The calculation gives us 3312−(33×38)=3312−1254=20583312 - (33 \times 38) = 3312 - 1254 = 2058. So the updated matrix is: ∣1amp;0amp;38 0amp;21amp;2058∣\begin{vmatrix} 1 & 0 & 38 \ 0 & 21 & 2058 \end{vmatrix} Next, we divide the second row by 2121 to solve for bb. Dividing 20582058 by 2121 gives us b=205821=98b = \frac{2058}{21} = 98. So, each share in the biotech firm costs \$(\$\(98\))\$.
  11. 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 \(3312 - (33 \times 38) = 3312 - 1254 = 2058\)\(0\).
  12. 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 = \(38\)\(3312 - (33 \times 38) = 3312 - 1254 = 2058\)\(2\).

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