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

Respuesta :

DeanR

[tex] x + y = 30[/tex]


[tex] 5x + 10 y = 210 [/tex]


As a matrix equation that's


[tex] \begin{pmatrix} 1 & 1 \\ 5 & 10 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix}30\\ 210 \end{pmatrix}[/tex]




Answer:

The equation could be

[tex]\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][/tex]

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

[tex]\left \{ {{x+y=30} \atop {0.05x+0.10y=2.10}} \right.[/tex]

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

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

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

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

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

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

This gives us the matrix equation above.