You can put this solution on YOUR website! If roger has 43 coins consisting of only dimes and quarters
d + q = 43
d = 43 - q; use this form for substitution
and in total he has 6.55$
.10d + .25q = 6.55
Replace d with (43-q)
.10(43 -q) + .25q = 6.55
4.3 - .10q + .25q = 6.55
.15q = 6.55 - 4.30
.15q = 2.25
q = 2.25/.15
q = 15 quarters
How many does he have of each coin
I'll let you find the no. of dimes, check the solutions using the 2nd equation