SOLUTION: Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $1

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $1      Log On


   

Question 155058: Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were 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?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Let n = # of nickels and d = # of dimes

Since "Joe has a collection of nickels and dimes that is worth $5.65", this means that the first equation is 0.05n%2B0.10d=5.65

0.05n%2B0.10d=5.65 Start with the first equation


5n%2B10d=565 Multiply every term by 100 to move the decimal point two places to the right (effectively eliminating the decimal)

Also, since "the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45", this tells us that 0.05%28n%2B8%29%2B0.10%282d%29=10.45

0.05%28n%29%2B0.05%288%29%2B0.10%282d%29=10.45 Distribute


0.05n%2B0.4%2B0.20d=10.45 Multiply


5n%2B40%2B20d=1045 Multiply every term by 100 to move the decimal point two places to the right to make the decimal numbers whole.


5n%2Bcross%2840-40%29%2B20d=1045-40 Subtract 40 from both sides


5n%2B20d=1005 Subtract. So this is our second equation

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So we have the system of equations:

system%285n%2B10d=565%2C5n%2B20d=1005%29


5n%2B10d=565 Start with the first equation


5n=565-10d Subtract 10d from both sides

Notice how there's a "5n" in the second equation. So simply replace the "5n" in the second equation with 565-10d

%28565-10d%29%2B20d=1005 Plug in 5n=565-10d


10d%2B565=1005 Combine like terms.


10d%2Bcross%28565-565%29=1005-565 Subtract 565 from both sides.


10d=440 Combine like terms


cross%2810%2F10%29d=440%2F10 Divide both sides by 10


d=44 Reduce


So Joe has 44 dimes.


5n%2B10d=565 Go back to the first equation


5n%2B10%2844%29=565 Plug in d=44


5n%2B440=565 Multiply


5n=125 Subtract 440 from both sides


n=25 Divide both sides by 5 to isolate "n"

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Answer:

So Joe has 25 nickels and 44 dimes