Solve the problem by setting up and solving a system of three linear equations in three variables.

Question

A box contains $8.15 in nickels, dimes, and quarters. There are 47 coins in all, and the sum of the numbers of nickels and dimes is 3 less than the number of quarters. How many coins of each kind are there?

Answer 1

We Have

1) N +D +Q = 47

2). .05N +.1D +.25Q = 8.15

and

3) N+D = Q-3 .

If we substutitue the right hand side of 3) into 1 we get

Q-3 -Q = 47 or 2Q = 44 divide both side by 2 and get Q=22.

we know 22 quarter are worth 5.50 substitute this into 2 and see

.05N + .1D +.5.50 = 8.15. or .05N +.1D = 2.65 Of we take this last expression and multiply both 20 we get N+ 2D=53. so we use N+2D= 53 and N+D+Q(=22 ) = 47

so N+2D= 53 and N+D = 25

If I subract 1 from the other I get

(N+2D)-(N+D)=D=53-25 = 28

Tis means N= -3. So this box contains 28 dimes, 22 quarter and an IOU for 3 nickels.

This was a trick question..

I hope this helps

Answer 2

Ali, my guess is that you’d like to know HOW to do it, so here is a wordier (and more accurate) demonstration.

First, always give labels for the quantities:

Let x be the number of nickels

Let y be the number of dimes

Let z be the number of wuarters

Then translate the other statements into math.

First, translate the total value of the coins,

recognizing that every nickels is worth 5 pennies, each fine is worth 10, and each quarter 25.

(a) 5x + 10y + 25 z = 815

Next translate the number of coins

(b) x + y + z = 47

Finally, translate the other sentence

(c) x + y = z - 3

(Notice Sam made a typo here; it will affect the outcome)

Now an algebra 2 student already knows how to solve three linear equations in three unknowns, and you may already know, too.

But for the early algebra student, I’ll continue the solution.

First, take equation (c) and move the z to the other side by subtracting z from both sides.

(d) x + y - z = -3

Now add (b) and (d), which gets rid of the z.

(e) 2x + 2y = 44

simplify by dividing by 2

(f) x + y = 22

Now discover z by subtracting (f) from (b)

(g) z = 25

Now let’s reduce the information in (a) by plugging in this value for z.

(h) 5x + 10y + 25* 25= 815

Collect terms and subtract

(I) 5x + 10y = 190

At this point we have two equations in two unknowns,

(f) and (I).

To solve these equations, solve (f) for x.

(j) x = 22 - y

and plug that into (I)

(k) 5(22-y) + 10y = 190

Multiply it all out

(L) 110-5y + 10y = 190

collect terms and solve for y

(m) 5y = 80

(n) y = 16

Finally, plug y and z into (c) to discover x

(p) x + 16 = 25 - 3

(q) x = 6

So there are 6 nickels, 16 dimes, and 25 quarters. Verify by plugging into (a)

6*5 + 16*10 + 25*25 ?=? 815

30 + 160 + 625 = 815 = 815

Answer 3

x5+y10+z25=815

x+y+z=47

x+y-3=z

2x+2y=50

x+y=25

12x+13y+22z=740?