SOLUTION: Hector has a collection of nickels, dimes, and quarters totaling 122 coins. The number of dimes he has is 3 more than four times the number of nickels, and the number of quarters

Question 68147This question is from textbook Algebra for College Students
: Hector has a collection of nickels, dimes, and quarters totaling 122 coins. The number of dimes he has is 3 more than four times the number of nickels, and the number of quarters he has is 19 less than the number of dimes. How many coins of each kind does he have? This question is from textbook Algebra for College Students

Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
Hector has a collection of nickels, dimes, and quarters totaling 122 coins. The number of dimes he has is 3 more than four times the number of nickels, and the number of quarters he has is 19 less than the number of dimes. How many coins of each kind does he have?
Let the number of Nickels be:x
Then the number of dimes is (3 more that 4 times nickels): 4x+3
Then the number of quarters is (19 less thatn the number of dimes: 4x+3-19=4x-16
The equation to solve is:
Nickels+dimes+quarters=122
(x)+(4x+3)+(4x-16)=122
9x-13=122
9x-13+13=122+13
9x+135
9x/9=135/9
x=15
:
Nickels: x=15
Dimes:4x+3=4(15)+3=63
Quarters: 4x-16=4(15)-16=44
:
Sanity check, do 15 nickels+63 dimes + 44 Quarters=122 coins?
15+63+44=122
122=122 Yes. I think we've got it!
Happy Calculating!!!
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