SOLUTION:
Please help me solve this question:
A jar contains 80 nickels and dimes worth $6.20. How many of each kind of coin are in the jar x=nickels y = dimes
Thank you
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Please help me solve this question:
A jar contains 80 nickels and dimes worth $6.20. How many of each kind of coin are in the jar x=nickels y = dimes
Thank you
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Question 852087:
Please help me solve this question:
A jar contains 80 nickels and dimes worth $6.20. How many of each kind of coin are in the jar x=nickels y = dimes
Thank you Answer by checkley79(3341) (Show Source):
You can put this solution on YOUR website! X=NICKELS.
Y=80-X DIMES.
.05X+.10Y=6.20
.05X+.10(80-X=6.20
.05X+8-.10X=6.20
-.05X=6.20-8.00
-.05X=-1.80
X=-1.80/.05
X=36 NICKELS.
Y=80-36=44 DIMES.
PROOF:
.05*36+.10*44=6.20
1.80+4.40=6.20
6.20=6.20
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