Question 141777: Sam emptied his collection of nickels and dimes on his bed. If his 46 coins were worth $3.10, how many coins of each type did he have? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=number of nickels
And let y=number of dimes
Now we are told:
x+y=46-------------------------------------------eq1
and($ understood)
0.05x+0.25y=3.10-------------------------------------eq2
Multiply eq1 by 0.05
0.05x+0.05y=2.3------------------------eq1a
Subtract eq1a from eq2
0.20y=0.80 divide both sides by 0.20
y=4--------------------------------------number of quarters
Substitute y=4 into eq1
x+4=46 subtract 4 from each side
x+4-4=46-4 collect like terms
x=42-----------------------------number of nickels
CK
42+4=46
46=46
and
4*0.25+42*0.05=3.10
1.00+2.10=3.10
3.10=3.10
Another way to solve this problem using 1 unknown:
Let x=number of nickels
Then 46-x=number of quarters
0.05x+0.25(46-x)=3.10 get rid of parens
0.05x+11.50-0.25x=3.10 subtract 11.50 from each side
0.05x+11.50-11.50-0.25x=3.10-11.50 collect like terms
-0.20x=-8.40 divide both sides by -0.20
x=42--------------------------------------------number of nickels
46-42=4--------------------------------------number of quarters
