Question

A man has 30 coins, which are all dimes or nickels. Their combined value is $2.10. How many coins of each kind does he have? Let x = the number of nickels. Let y = the number of dimes. Which of the following matrices could be used to solve the problem

Answer 1

Answer:

The equation could be

$\left[\begin{array}{ccc}1&1\\0.05&0.1\end{array}\right] \times \left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}30\\2.10\end{array}\right]$

Step-by-step explanation:

The equation to represent the number of coins would be

x + y = 30

The equation to represent the amount of money the coins are worth would be

0.05x + 0.10y = 2.10

This gives us the system

$\left \{ {{x+y=30} \atop {0.05x+0.10y=2.10}} \right.$

Using matrices, the first matrix will be the coefficient matrix.  This contains the coefficients for x and y for both equations:

$\left[\begin{array}{ccc}1&1\\0.05&0.10\end{array}\right]$

The second matrix will be the variable matrix, containing the variables for both equations:

$\left[\begin{array}{ccc}x\\y\end{array}\right]$

The matrix after the equals sign will be the constant matrix:

$\left[\begin{array}{ccc}30\\2.10\end{array}\right]$

This gives us the matrix equation above.

Answer 2

$x + y = 30$

$5x + 10 y = 210$

As a matrix equation that's

$\begin{pmatrix} 1 & 1 \\ 5 & 10 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix}30\\ 210 \end{pmatrix}$