Question 1145667
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Using formal algebra....<br>
d = # of dimes
q = # of quarters<br>
(1) 10d+25q = 620  [the value of the dimes (10 cents each) and quarters (25 cents each) is $6.20 = 620 cents]
(2) d = 3q+18  [the number of dimes is 18 more than 3 times the number of quarters]<br>
Solve the pair of equations.  Probably the easiest method is to substitute (2) into (1) and solve for q; then use that value of q in either equation to solve for d.<br>
I leave it to you to get the practice finishing solving the problem using formal algebra.<br>
Using logical reasoning and simple mental arithmetic....<br>
(1) Count the 18 "extra" dimes.  Their value is 180 cents; that leaves 440 cents for the remaining coins.
(2) The remaining coins are 3 dimes for each quarter.  Imagine the remaining coins in groups each consisting of 1 quarter and 3 dimes.  The value of 1 quarter and 3 dimes is 55 cents.
(3) The number of groups at 55 cents each required to make the remaining 440 cents is 440/55 = 8.  In those 8 groups there are 8 quarters and 8*3=24 dimes.
(4) So all together there are 8 quarters and 24+18=42 dimes.<br>
 
(3) The number of groups of