Question 204021
Well, the first step is to assign variables to each of the unknowns in the problem:
Let:
H = # of half-dollars.
Q = # of quarters.
D = # of dimes.
N = # of nickels.
Now to translate the English prose to algebraic statements:
Q = 2H "...twice as many quarters as half-dollars,..."
D = 2Q "...twice as many dimes as quarters,..."
N = 3D "...three times as many nickels as dimes..."
Now we can write an equation relating all four varables (H, Q, D, and N) to the total amount tendered at the bank ($10.00)
For example, the dollar-value of H half dollars can be expressed as: (0.5)H, so...
(0.5)H+(0.25)Q+(0.1)D+(0.05)N = $10.00
Now the trick is to reduce the number of variables from four down to one! It really doesn't matter which variable you choose but try to pick the easiest path to getting it.
I picked the variable H, the number of half-dollars.
Here we go!
N = 3D substitute D = 2Q to get:
N = 3(2Q) = 6Q substitute Q = 2H
N = 3(2(2H)) or (3(4H)) so...
N = 12H
D = 2Q substitute Q = 2H, so...
D = 2(2H) = 4H
So now we have:
Q = 2H
D = 4H
N = 12H so we can substitute into the first equation to get:
(0.5)H+(0.25)(2H)+(0.1)(4H)+(0.05)(12H) = 10 Add the H's together:
(0.5)H+(0.5)H+0.4H+0.6H = 10
2H = 10 Divide both sides by 2.
H = 5 and...
Q = 2H = 10 and...
D = 4H = 20 and...
N = 12H = 60, so...
The store-keeper got:
5 Half-dollars.
10 Quarters.
20 Dimes.
60 Nickels.
Now let's check to see if this all adds up tp $10.00
(0.5)5+(0.25)(10)+(0.1)(20)+(0.05)(60) = 2.5+2.5+2+3 = 10 OK!