# Algebra I

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lauren has 40 coins all together, all dimes, nickles, and quarters they equal \$4.50 and he has 7 more dimes than nickles, how many quarters does he have

• Algebra I -

Let n = number of nickels.
Let d = number of dimes.
Let q = number of quaraters.

n + d + q = ____

5n + 10d + 25q = 450 (Why?)

d = n + 7 (Why?)

Rewrite the first to equations but use (n + 7) everywhere you see d.

n + (n + 7) + q = ???
5n + 10(n + 7) + 25q = 450

You have 2 equations with 2 variables and you should now be able to solve using methods you have learned for solving a "system" of equations. Or you could try graphing the lines that the equations represent and see where the lines cross.

Another way is to think about all the ways that you could have 40 coins and yet have 7 more dimes than nickels:

n d q
0 7 33
1 8 31
2 9 29
3 10 27
4 11 25
5 12 23

etc

You could calculate the value of each set of coins until you find the one that has a value of \$4.50. That would be the 'brute force' method! I'm guessing you don't really have that many quarters, so it might be better to start looking later in the list.

Another thing to think about: the list shows that number of quarters must be odd. So total value from quarters will end in 5 cents (.25, .75, 1.25, etc.). No matter how many dimes you have, their value will end in zero (.10, .20, .30), so to get a value of \$4.50, you will need to have an odd number of nickels to 'cancel out' the 5 cents in the value of the quarters.

So I'm guessing the answer will have an odd number of nickels, an odd number of quarters, and an even number of dimes. That also helps you cut down the list.

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