Question 669400

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?    

This is the problem directly from my homework.  I'm trying to figure out the system of equations so I can solve the problem.  So far I've come up with

	.1d+.05n=5.65       and               2(.1d)+.05n+8=10.45

but I keep ending up with d= -32 so I must be way off.


You are off, but not by that much. Looking at what I got and you got, I can see that your 2nd equation should have been .1(2D) + .05(N + 8) = 10.45. This is where your error is. The correct solution is below:


Let the original amount of dimes and nickels that he has be D, and N, respectively
Then: .1D + .05N = 5.65 ----- eq (i)
Doubling the amount of dimes would make the dimes, 2D
Increasing the nickels by 8 would make the amount of nickels, N + 8 
Therefore, .1(2D) + .05(N + 8) = 10.45 ------ .2D + .05N + .4 = 10.45 ---- .2D + .05N = 10.05 ----- eq (ii)


.1D + .05N = 5.65 ---- eq (i)
.2D + .05N = 10.05 ---- eq (ii) 
- .1D = - 4.40 --------- Subtracting eq (ii) from eq (i)
D, or original amount of dimes = {{{(- 4.4)/- .1}}}, or {{{highlight_green(44)}}}


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