Asked by tdfg
Sue had $5.65 worth of quarters and dimes. There were 3 more quarters than dimes. How many quarters were there?
Answers
Answered by
oobleck
10(q-3) + 25q = 565
so solve for q
so solve for q
Answered by
PsyDAG
d = Q - 3
.25Q + .10d = .25Q + .10(Q-3) = $5.65
Solve for Q.
.25Q + .10d = .25Q + .10(Q-3) = $5.65
Solve for Q.
Answered by
CookieCrumbs
number of dimes = y
number of quarters = y+3
10y + 25(y+3) = 565
solve for y
number of quarters = y+3
10y + 25(y+3) = 565
solve for y
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