Question 687690: Diane has $1.30 in dimes and nickels. She has a total of 16 coins. How many of each kind does she have?
Answer by sp75(12)   (Show Source):
You can put this solution on YOUR website! To solve this problem, you need to create a system of equations (a set of equations that work with the same set of variables). Why? Well, you have two variables and no equations. If you want to solve for those variables, you will need at least two equations using the same two variable.
Now that the intro is over, let's zone in on the problem.
1. What do we know?
-Diane has $1.30
-she ONLY has dimes and nickels
----> Why is it so important that they specify dimes and nickels?
They want us to take note of their monetary value(.10 cents and .05 cents)
-She has a total of 16 coins
---->Why is this important?
It tells us that no matter how many coins she has, the number of dimes plus
the number of nickels will equal 16. Wait! That sounds like an equation in
the making!
Let's say we call the number of dimes "d" and the number of nickels "n".
That means d + n = 16.
Just by looking at the information, we built one equation. We just need one
more.
2. Find the next equation
Well, the other information we have gave us an equation based off of the quantity of coins Diane has. The only other thing we know is that everything is equal to $1.30 --- monetary value. But is that really all we know? Let's see.
Do you remember talking about the monetary value of dimes and nickels?
What if we said that the monetary value of the numbers of dimes we have plus the
monetary value of the number of nickels we have is equal to our total monetary value.
So.... .10*d + .05*n = $1.30 . This is our second equation.
3. Set up and solve the system for each variable (d and n)
.10*d + .05*n = $1.30
+ d + n = 16
At this point, we need to solve for one variable. We can do this by getting rid of one. All you need to choose the variable you want to get rid of, in one equation make sure it is positive and in the other, negative. The coefficient of that variable (the number in front) also has to equal to that in the other.
This is one of those problems where an example seems necessary.
 <--I moved each decimal to the right two places to make
 the algebra easier
Now I am going to multiply the top by -1/5 and I get...
Do you see? Now, when I add both of the equations together, The n's will cancel out, and I will be able to solve for d.
-d = -10
d = 10 <----So you have ten dimes
Now plug in d into either one of the equations (it doesn't matter which), and solve for n.
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