Hi Kenny,
This problem requires you to create a 'system of equations'. Because you have two unknowns (dimes and nickels) we are going to need two equations; one that deals with the number of coins, and one that deals with the value of the coins.
The first equation is:
d + n = 71 (the number of dimes and nickels all together is 71);
The second equation is:
.10d + .05n = 4.75 (each dime is worth .10 cents and each nickel is worth .05cents and together, they add up to $4.75)
I don't like working with decimals, so I'm going to multiply each term in the above equations by 100 so that I can work with whole numbers. The new equation is
10d + 5n = 475
Next, I'm going back to my first equations and solve for d in terms of n (just subtract n from both sides) and we are left with
d = 71 - n. Now, I'm going to substitute this 71 - n for d in the second equation.
10(71 - n) + 5n = 475. Yeah! This is going to be easy because now I have one equation and one variable. I just need to solve for n.
710 - 10n + 5n = 475 (combine like terms)
710 - 5n = 475
-5n = -235 (subtract 710 from both sides)
n = 47 (divide both sides by the coefficient of the variable -5)
So, there are 47 nickels.
back to our original equation
d + n = 71 We now know that there are 47 nickels so
d + 47 = 71 therefore
d = 24 . There are 24 dimes.
24 dimes and 47 nickels.
Hope that helps!
