SOLUTION: Okay, So, I have $7.30 in nickels, dimes & quarters. The number of dimes is 2 less than twice the number of nickels, and the number of quarters is 1 more than three times the numbe

Question 1039121: Okay, So, I have $7.30 in nickels, dimes & quarters. The number of dimes is 2 less than twice the number of nickels, and the number of quarters is 1 more than three times the number or nickels. How many of each should I have?
I made nickels my x. This is how I wrote it out. 5x+20x-20+75x+25= 730. I ended up with 100x-5= 730, dividing the right and left side by 100 and ended up with 7.25. I keep coming up with this, two of my classmates did, too. We are at loss as how to divide into the correct coins. Where did we go wrong? We have our test on Monday.
Thank you for any help you can provide.
Rochelle
Found 2 solutions by addingup, MathTherapy:
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
0.05n+0.10d+0.25q = 7.30
d = 2n-2
q = 3n+1
Substitute in the first equation:
0.05n+0.1(2n-2)+0.25(2n+1)= 7.30

Answer by MathTherapy(10557) (Show Source): You can put this solution on YOUR website!

Okay, So, I have $7.30 in nickels, dimes & quarters. The number of dimes is 2 less than twice the number of nickels, and the number of quarters is 1 more than three times the number or nickels. How many of each should I have?
I made nickels my x. This is how I wrote it out. 5x+20x-20+75x+25= 730. I ended up with 100x-5= 730, dividing the right and left side by 100 and ended up with 7.25. I keep coming up with this, two of my classmates did, too. We are at loss as how to divide into the correct coins. Where did we go wrong? We have our test on Monday.
Thank you for any help you can provide.
Rochelle

I don't see you going wrong anywhere. Check the problem again. I have a feeling that something was not copied right.
I also got: 5x + 20x � 20 + 75x + 25 = 730, or 100x = 725, which makes x a non-integer, but it MUST be an integer.
Again, check the problem.
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