Question 29464: I have tried this coin problem several ways and can't get it to add up right. Supposidly we are to answer it using the elimination method. The question is.
Sam has a collection of nickels and dimes that is worth $5.65. If the number of dimes wer doubled and the number of nickels were increased by 8. The value of the coins would be $10.45. How many dimes does he have?
I set it up like this. x= nickles y = dimes
5x + 10y = 5.65
5(x + 8) + 2(10y) = 10.45
I came up with x = 9, and y = 44. There is no way that is right. I have spent hours on this. Please help!!
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE MY COMMENTS BELOW..
I have tried this coin problem several ways and can't get it to add up right. Supposidly we are to answer it using the elimination method. The question is.
Sam has a collection of nickels and dimes that is worth $5.65. If the number of dimes wer doubled and the number of nickels were increased by 8. The value of the coins would be $10.45. How many dimes does he have?
I set it up like this. x= nickles y = dimes................GOOD
5x + 10y = 5.65............IT SHOULD BE 565 CENTS...NOT 5.65 $..SINCE ON LHS,YOU ARE USING 5 AND 10 AS CENTS..YOU SHOULD PUT 0.05 AND 0.10 IF YOU WANT TO USE 5.65 $..CONSISTENCY OF UNITS...
5(x + 8) + 2(10y) = 10.45...SAME COMMENT ..IT IS 1045 CENTS.ALSO BETTER WRITE 10(2Y)...TO SHOW THAT THE DIMES ARE DOUBLED,AS YOU HAVE PUT 5(X+8)TO SHOW THAT NICKELS ARE INCREASED BY 8.
I came up with x = 9, and y = 44. There is no way that is right. I have spent hours on this. Please help!!
SO WE HAVE 5X+10Y=565...OR DIVIDING BY 5 THROUGHOUT...
X+2Y=113........................I
5X+40+20Y=1045...OR.....5X+20Y=1045-40=1005......DIVIDING WITH 5...
X+4Y=201....................II
EQNII-EQN.I
X+4Y-X-2Y=201-113=88
2Y=88...OR Y=44................YOU GOT IT OK!!!!
SUBSTITUTING IN EQN.I....
X+2*44=113
X=113-88=25.....YOU GOT IT WRONG HERE!!...HOW?CHECK BACK
SO X=25 AND Y=44
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