SOLUTION: Joe has a collection of nickles and dimes that is worth, $7.05. If the # of dimes were doubled, and the # of nicles were increased by eight, the value of the coins would be, $11.75

Question 172826: Joe has a collection of nickles and dimes that is worth, $7.05. If the # of dimes were doubled, and the # of nicles were increased by eight, the value of the coins would be, $11.75. How many dimes does he have?
It really helps me to understand if I can See each step to solve this problem written out like in a math book. Thank you
Found 3 solutions by checkley77, solver91311, josmiceli:
Answer by checkley77(12844)   (Show Source): You can put this solution on YOUR website!
Let D=the number of dimes.
Let N=the number of nickles.
.10D+.05N=7.05 (multiply this equation by -2 & add to the other equation
.10*2D+.05(N+8)=11.75
.20D+.05N+.40=11.75
.20D+.05N=11.75-.40
.20D+.05N=11.35
-.20D-.10N=-14.10
-----------------------
-.05N=-2.75
N=-2.75/-.05
N=55 nickles.
.10D+.05*55=7.05
.10D+2.75=7.05
.10D=7.05-2.75
.10D=4.30
D=4.30/.10
D=43 dimes.
Proof:
.20*43+.05*55=11.35
8.60+2.75=11.35
11.35=11.35

Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!
Let be the number of dimes and be the number of nickels.

That means that the value of his dimes is cents and the value of his nickels is cents.

I expressed those values in terms of cents because it is less messy if you don't have to deal with decimal fractions as coefficients in the equations, but the problem would have worked out just as well to say the value of the dimes is dollars. The only thing you have to remember if you express the value in cents is to change the total value numbers, i.e. $7.05 and $11.75 to 705 cents and 1175 cents.

The first thing we know is that

Next, if we double the number of dimes, we now have dimes, the value of which is cents. If we increase the number of nickels by 8, then the value of the nickels becomes cents.

Now we can say . Collect like terms and put it in standard form and you get .

That gives us a system of two equations in two variables that can be solved by a variety of methods. I choose to use the elimination method because I can see that the coefficients in the two equations lend themselves to a tidy solution by using elimination.

Step 1: Multiply by to get

Step 2: Add the result of Step 1 to the second equation, namely to get or, after simplifying, (Notice how the coefficient on went to zero, effectively eliminating that variable? Hence the name.)

Step 3: Solve for

telling us that the original number of dimes was 43.

But you aren't quite done yet, even though you have answered the question. We need to check our work to make sure we didn't make some silly mistake.

If the original number of dimes was 43, then there must have been $4.30 in dimes. $7.05 minus $4.30 is $2.75, meaning that there must have been $2.75 worth of nickels. Dividing by $0.05 we get 55, telling us there were 55 nickels to begin with.
If we double the dimes, we get 2 times 43 equals 86 dimes meaning we now have $8.60 worth of dimes, and if we add 8 nickels we get $2.75 plus $0.40 equals $3.15 worth of nickels.

$8.60 plus $3.15 equals $11.75 which was the amount given in the problem. Answer checks. NOW you are done.
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
I always start a problem like this by replacing the things
I'm looking for with letters
Let = number of nickels Joe has
Let = number of dimes Joe has
Now I put in equation form the information I'm given
(1)
Each term of this equation represents cents. In words:
(5 cents/nickel x # of nickels) + (10 cents/dime x # of dimes) = 705 cents
Or,if I wrote it the following way, each term represents dollars
, but it's the same equation
Also given:
(2)
Now I have 2 equations and 2 unknowns, so that's all I need to solve it
(2) needs some work
(2)

Now subtract (1) from this

Joe has 43 dimes answer
Now substitute this in (1) to find
(1)

Now I can check my answer:
(2)

OK
And also
(1)

OK

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