Question

A shipping company charges a flat rate of $7 for packages weighing five pounds or less, $15 for packages weighing more than five pounds but less than ten pounds, and $22 for packages weighing more than ten pounds. During one hour, the company had 13 packages that totaled $168. The number of packages weighing five pounds or less was three more than those weighing more than ten pounds. The system of equations below represents the situation.
s+m+l=13
7s+15m+22l=168
s-l=3

Which matrix can be used to show the number of packages in the different weight classes during that hour?

Answer 1

Answer:

The matrix equation would be

$\left[\begin{array}{ccc}1&1&1\\7&15&22\\1&0&-1\end{array}\right] \times \left[\begin{array}{ccc}s\\m\\l\end{array}\right] =\left[\begin{array}{ccc}13\\168\\3\end{array}\right]$,

and the resulting matrix is

$\left[\begin{array}{ccc}6\\4\\3\end{array}\right]$

Step-by-step explanation:

We first start with what is called the coefficient matrix; it contains the coefficients of the variables in the system of equations.

The coefficients of the first equation are 1, 1 and 1; this will be the first row of the matrix.

The coefficients of the second equation are 7, 15 and 22; this will be the second row of the matrix.

The coefficients of the third equation are 1, 0 and -1 (0 because there is no variable m in the equation); this will be the third row of the matrix, and gives us the coefficient matrix

$\left[\begin{array}{ccc}1&1&1\\7&15&22\\1&0&-1\end{array}\right]$

The next matrix in the equation will be the variable matrix.  It will contain the variables for the equation; for ours, it is s, m and l:

$\left[\begin{array}{ccc}s\\m\\l\end{array}\right]$

The last matrix in our equation will be the constant matrix, which has the values after the equals in each equation:

$\left[\begin{array}{ccc}13\\168\\3\end{array}\right]$

This gives us the equation:

$\left[\begin{array}{ccc}1&1&1\\7&15&22\\1&0&-1\end{array}\right] \times \left[\begin{array}{ccc}s\\m\\l\end{array}\right] =\left[\begin{array}{ccc}13\\168\\3\end{array}\right]$

To solve this, we find the inverse of the coefficient matrix and multiply it by the constant matrix:

$\left[\begin{array}{ccc}s\\m\\l\end{array}\right] =\left[\begin{array}{ccc}1&1&1\\7&15&22\\1&0&-1\end{array}\right] ^{-1} \times \left[\begin{array}{ccc}13\\168\\3\end{array}\right]$

Using a graphing calculator to perform this, we get the result

$\left[\begin{array}{ccc}6\\4\\3\end{array}\right]$

This means the small contains 6 packages; medium contains 4; and large contains 3.

Answer 2

Should be the first one with elcholen reducing form shown, and sides going 6, 4, 3. Showing packages and them being multiplied by its value of each box to eqaul 168. 6 = s 4 = m 3 = L