SOLUTION: I need some help with this word problem, setting it up to solve. Thanks. Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: I need some help with this word problem, setting it up to solve. Thanks. Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and       Log On


   



Question 143228: I need some help with this word problem, setting it up to solve. Thanks.
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?

Found 2 solutions by vleith, solver91311:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Let d be the number of dimes, n be the number of nickels
Given: 10d+%2B+5n+=+565
also given: 10%282d%29+%2B+5%28n%2B8%29+=+1045
10%282d%29+%2B+5n+%2B+40+=+1045
20d+%2B+5n++=+1005
Solve for d
10d+%2B+5n+=+565
20d+%2B+5n++=+1005 use elimination and subtract
-----------------------
-10d++++++=+-440
+d+=+44
check your answer. if d+=+44, then n+=+%28565-440%29%2F5= 25
Verify using the second equation.
20d+%2B+5%28n%2B8%29+=+1045
20%2A44+%2B+5%2825%2B8%29+=+1045
880+%2B+5%2A33+=+1045
880+%2B+165+=+1045
check!

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Number of dimes: d
Number of nickels: n
Value (in cents) of d dimes: 10d
Value (in cents) of n nickels: 5n
Value (in cents) of current collection: 565
Number of dimes in the 'what if' collection: 2d
Number of nickels in the 'what if' collection: n+%2B+8
Value (in cents) of the dimes in the 'what if' collection: 10%282d%29
Value (in cents) of the nickels in the 'what if' collection: 5%28n%2B8%29
Value (in cents) of 'what if' collection: 1045

10d%2B5n=565 and
10%282d%29%2B5%28n%2B8%29=1045 => 20d%2B5n%2B40=1045 => 20d%2B5n=1005

Now you have 2 equations in 2 variables. Solve for d.

I changed everything from dollars/cents to cents just because I prefer integer coefficients as opposed to decimal fraction coefficients.