I have 56 coins consisting of nickels and quarters. The total value of the coins is $5.60. How many of each coin do I have?

Algebra to the rescue...

NN is number of nickles
ND is the number of quarters.

NN*.05 + ND*.25=$5.60
but NN + ND=56

two equations, two unknowns, solve.

its very simple.
let one varible be a representation of the other one.
from there you can plug the equation in.

whenever you have two equations and two variables in each, you follow the steps above to solve them.

How does 42 nickels 14 quarters sound?

n:number of nickels
q:number of quarters

Equation 1: n+q=56
Equation 2: 5n+25q=560

multiply all of Equation 1 by 5-->
5n+5q=280
then subtract this new equation from Equation 2
5n+25q=560
5n+5q=280
-_________
20q=280
solve for q
q=14
therefore n=42

To solve this problem, we will use algebraic equations. Let's assign variables to the unknowns:

Let n be the number of nickels and q be the number of quarters.

Since the total number of coins is 56, we can create the equation: n + q = 56.

The value of each nickel is $0.05, so the total value of nickels is 0.05n.
Similarly, the value of each quarter is $0.25, so the total value of quarters is 0.25q.

The total value of the coins is given as $5.60, so we can create another equation:

0.05n + 0.25q = 5.60.

Now we have a system of two equations with two unknowns:
Equation 1: n + q = 56
Equation 2: 0.05n + 0.25q = 5.60.

To solve this system, we can use the method of substitution or elimination.

Let's use substitution:
From Equation 1, we can isolate n: n = 56 - q.

Substituting this into Equation 2, we get: 0.05(56 - q) + 0.25q = 5.60.

Simplifying, we have: 2.8 - 0.05q + 0.25q = 5.60.

Combining like terms, we get: 0.20q + 2.8 = 5.60.

Subtracting 2.8 from both sides, we have: 0.20q = 2.80.

Dividing both sides by 0.20, we get: q = 14.

Now that we have the value for q, we can substitute it back into Equation 1 to find n:

n + 14 = 56.
Subtracting 14 from both sides, we get: n = 42.

Therefore, you have 42 nickels and 14 quarters.