Solving a system with 3 variables

Question

A collection of nickles, dimes, and quarters consists of 21 coins with a total of $2.45. If the number of dimes equals the number of nickles, find the number of each type of coin

Answer 1

Let N = the number of nickels, D = the number od dimes, and Q = the number of quarters.

Nickles, dimes, and quarters consists of 21 coins means N + D + Q = 21

A total value of $2.45 means .05N + 0.10D + 0.25Q = 2.45 or (multiplying by 40) 2N + 4D + 10Q = 98

The number of dimes equaling the number of nickles means D = N or N - D = 0

So the three equations you have are:

N + D + Q = 21

2N + 4D + 10Q = 98

N - D = 0

I like to use Elimination as my method of choice. So, I would first choose 1 variable to eliminate. I'll choose Q because I already have one equation without a Q in it. So to eliminate Q from equations 1 & 2, I would multiply equation 1 by -10 to get: -10N - 10D -10Q = -210 and add it to equation 2:

-10N - 10D -10Q = -210

2N + 4D + 10Q = 98

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-8N - 6D = -112

Now I have 2 equations in 2 unknowns:

-8N - 6D = -112

N - D = 0

Now multiply N - D = 0 by 8 to get 8N - 8D = 0 and add the equations:

-8N - 6D = -112

8N - 8D = 0

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-14D = -112

or D = 8

Since N = D, then N = 8

Now, plug those into one of the original equations to find Q. I'll pick N + D + Q = 21 so 8 + 8 + Q = 21 or Q = 5.

Now check by using the other equation:

.05N + 0.10D + 0.25Q = 2.45

.05(8) + 0.10(8) + 0.25(5) = 2.45

.40 + .80 + 1.25 = 2.45

2.45 = 2.45 We did it!!

Answer 2

Let N = number of nickels

Let D = number of dimes

Let Q = number of quarters

The total number of coins is comes from the quantity of coins (counting the coins). So...

N + D + Q = 21

The total value of the coins is $ 2.45. We know each nickel is 5 cents (.05) of the value, each dime is ten cents( .10) of the value and each quarter is 25 cents (.25) of the value.

.05 N + .10 D + .25 Q = 2.45

And finally it tells us number of dimes = number of nickels so...

D = N

we know have a system of three equations with three variables

EQ1 N + D + Q = 21

EQ 2 05 N + .10 D + .25 Q = 2.45

EQ 3 D = N

looking at EQ3 We can use substitution to replace the D's with N's in EQ1 and EQ2. So we now have a system of 2 equations with twp variables that can be solved using substituion.

N + N + Q =21

.05 N + .10 N + .25 Q = 2.45

2N + Q = 21

.15 N + .25 Q = 2.45

Once you solve the system above for N and Q , you can plug into EQ 1 of original system to find D

Answer 3

n represents the number of nickels

n represents the number of dimes

21 - 2n represents the number of quarters

0.10n + 0.05n + 0.25(21 - 2n) = 2.45

Can you solve for n and answer?