Answer:
Becky has 8 nickels, 6 dimes and 5 quarters in her pocket.
Explanation:
Let Number of nickels be 'n'
Let Number of dimes be 'd'.
Also Number of quarters be 'q'.
Given: there are two more nickels than dimes
hence
![n = 2+d](https://img.qammunity.org/2021/formulas/mathematics/middle-school/kt4zom584eewx94efyaozz4srpfpk044wn.png)
Now Given:
Total Number of Coins = 19
So the equation can be framed as;
![n+d+q=19\\\\2+d+d+q=19\\\\2d+q=19-2\\\\2d+q=17 \ \ \ \ equation \ 1](https://img.qammunity.org/2021/formulas/mathematics/middle-school/2eyhyb0jmj2jlcg9shptsd3stiorxnq140.png)
Also Given:
Total Amount in pocket = $2.25
Now we know that;
1 nickel = $0.05
1 dime = $0.1
1 quarter =$0.25
So the equation can be framed as;
![0.05n+0.1d+0.25q = 2.25\\\\0.05(2+d)+0.1d+0.25q=2.25\\\\0.1+0.05d+0.1d+0.25q=2.25\\\\0.15d+0.25q=2.25-0.1\\\\0.15d+0.25q=2.15 \ \ \ \ equation\ 2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/bvm4nl667g230tyyiuy7atbda9v65eypgy.png)
Now Multiplying equation 2 by 4 we get;
![4(0.15d+0.25q)=2.15*4\\\\4*0.15d+4*0.25q=8.6\\\\0.6d+q=8.6 \ \ \ \ equation \ 3](https://img.qammunity.org/2021/formulas/mathematics/middle-school/oslgjjieeyqueienflc217pawwf3k109oi.png)
Now Subtracting equation 3 from equation 1 we get;
Now Substituting the value of d in equation 1 we get;
![2d+q=17\\\\2*6+q=17\\\\12+q=17\\\\q=17-12\\\\q=5](https://img.qammunity.org/2021/formulas/mathematics/middle-school/btlm7a06jo1u0s7u2ot5imxc2ujjcx4b9z.png)
Also;
![n=2+d =2+6 =8](https://img.qammunity.org/2021/formulas/mathematics/middle-school/n7wvuvh09e101m3owlaihksu3kvltjzljd.png)
Hence Becky has 8 nickels, 6 dimes and 5 quarters in her pocket.