Solution:
Given:
Let d represent the number of dimes coins.
Let n represent the number of nickels coins.
Recall:
![\begin{gathered} 1dime=10cents=\text{ \$0.10} \\ 1nickel=5cents=\text{ \$0.05} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zrfhu20cfcxpsl0zacf7cy2edcpbunfaob.png)
To generate the system of equations:
![\begin{gathered} She\text{ has a total of 50 coins:} \\ d+n=50..................................(1) \\ \\ \\ Total\text{ value of the 50 coins:} \\ 0.1d+0.05n=4.50.......................(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cdna564rakokrs2lks8tth626efcp7wzpi.png)
From equation (1);
![\begin{gathered} d+n=50 \\ d=50-n................................(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hqv8skji2u23q9ltimctkve3pxrhig9n8q.png)
Substitute equation (3) into equation (2);
![\begin{gathered} 0.1d+0.05n=4.50 \\ 0.1(50-n)+0.05n=4.5 \\ 5-0.1n+0.05n=4.5 \\ 5-0.05n=4.5 \\ 5-4.5=0.05n \\ 0.5=0.05n \\ Divide\text{ both sides by 0.05 to get }n \\ (0.5)/(0.05)=n \\ n=10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/h4ot0c8q9oz2riv7vwgqaffwwqc402m36q.png)
Substitute the value of n into equation (3);
![\begin{gathered} d=50-n \\ d=50-10 \\ d=40 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/oq4xumdahgupsq2j0qpmrw90h1jjewbbcs.png)
Therefore, Daphne has 40 dimes coins and 10 nickels coins.