There are 11 nickels and 17 dimes
<em><u>Solution:</u></em>
Let "d" be the number of dimes
Let "n" be the number of nickels
Remember that a dime is worth 10 cents and nickel is worth 5 cents
<em><u>He has 28 coins worth $2.25</u></em>
number of dimes + number of nickels = 28
d + n = 28 --------- eqn 1
<em><u>The worth is $ 2.25. Therefore, we frame a equation as:</u></em>
2.25 dollar is worth 225 cents
number of dimes x Value of 1 dime + number of nickels x Value of 1 nickel = 225
From eqn 1,
d = 28 - n -------- eqn 3
<em><u>Substitute eqn 3 in eqn 2</u></em>
10(28 - n) + 5n = 225
280 - 10n + 5n = 225
5n = 55
<h3>n = 11</h3>
<em><u>Substitute n = 11 in eqn 3</u></em>
d = 28 - 11
<h3>d = 17</h3>
Thus there are 11 nickels and 17 dimes