<h3>30 dimes and 21 quarters are present</h3>
<em><u>Solution:</u></em>
Let "d" be the number of dimes
Let "q" be the number of quarters
1 dime = $ 0.10
1 quarter = $ 0.25
The worker counts 51 coins
Therefore,
d + q = 51
d = 51 - q ---------- eqn 1
<em><u>A parking meter contains $8.25 in dimes and quarters</u></em>
Therefore,
0.10d + 0.25q = 8.25 -------- eqn 2
<em><u>Substitute eqn 1 in eqn 2</u></em>
0.10(51 - q) + 0.25q = 8.25
5.1 - 0.10q + 0.25q = 8.25
0.15q = 3.15
Divide both sides by 0.15
<h3>q = 21</h3>
Substitute q = 21 in eqn 1
d = 51 - 21
<h3>d = 30</h3>
Thus 30 dimes and 21 quarters are present
which becomes 
and you get