math question?

Question

Miguel has 14 coins in his pocket. He has one more dime than quarters and three more nickels than dimes. How many of each coin does he have?

Okay, I know this is supposed to be a simple problem...I have the answer:

3 quarters
4 dimes
7 nickels

But how do you set it up? help please

Answer 1

He has a total of 14 coins (# quarters + # dimes + # nickels):

q + d + n = 14

He has one more dime than quarters (i.e. # dimes = # quarters + 1):

d = q + 1

He has three more nickels than dimes (i.e. # nickels = # dimes + 3):

n = d + 3

Use the second equation to substitute in (q + 1) for d in the third equation:

n = (q + 1) + 3 = q + 4

Then substitute it again for d in the first equation, and substitute (q + 4) for n, so that everything is in terms of the # of quarters:

q + (q + 1) + (q + 4) = 14

3q + 5 = 14

3q = 9

q = 3

Then plug back into your equations for dimes:

d = q + 1 = 3 + 1 = 4

And quarters:

n = q + 4 = 3 + 4 = 7

Answer 2

Let's call "q" the number of quarters, "d" the number of dimes, and "n" the number of nickels.

"Miguel has 14 coins in his pocket." So q + d + n = 14.

"He has one more dime than quarters..." So d = q + 1.

"...and three more nickels than dimes." So n = 3 + d.

So we have to solve the following system of equations:
q + d + n = 14
d = q + 1
n = 3 + d

Let's solve it by substitution. Let's put the first equation entirely in terms of q. Since d = q + 1, we can substitute q + 1 in for d in the first equation, and we get

q + (q + 1) + n = 14

Since n = 3 + d, and d = q + 1, then n = 3 + (q + 1), or n = q + 4. So we can substitute q + 4 in for n in the first equation, and we get

q + (q + 1) + (q + 4) = 14
3q + 5 = 14
3q = 9
q = 3.

Plugging q = 3 into the second equation and solving gives d = 4; plugging d = 4 into the third equation and solving gives n =7.

Answer 3

If q is the number of quarters, d the number of dimes, and n the number of nickels, then...

There is one more dime than there are quarters, so d = q + 1.
There are three more nickels than dimes, so n = d + 3.
Through substitution, n = (q + 1) + 3.

The rest is simply solving for q, given that q + d + n = 14:
q + (q + 1) + (q + 1 + 3) = 14
(q + q + q) + (1 + 1 + 3) = 14
3q + 5 = 14
3q = 9
q = 3

Then substitute 3 back for q:
d = q + 1 = (3) + 1 = 4
n = q + 1 + 3 = (3) + 4 = 7

Therefore, there are three quarters, four dimes, and seven nickels.

Answer 4

Let x be the number of quarters.
Therefore x + 1 will be the number of dimes and
x +1 +3 will be the number of nickels.

x (number of quarters) + (x +1 ) (number of dimes) + (x +1 +3) (number of nickels) = 14

x +x +1 +x+1+3 = 14
3x + 5 = 14
3x = 9
x = 3

Therefore, there are 3 quarters, 4 dimes and 7 nickels.

Answer 5

Let x = the number of quarters
Then x+1 = the number of dimes
So x+1 +3 =x+4 = number of nickles
So x + x+1 +x+4 + 14
3x + 5 = 14
3x= 9
x = 3 = number of quarters
x+1 = 4 =number of dimes
x=4 = 7 = nunber of nickles

Answer 6

He has the fewest quarters, so let number of quarters be q.

Dimes = q + 1
Nickels = q + 1 + 3 = q + 4

q + q + 1 + q + 4 = 14
3q + 5 = 14
3q = 9
q = 3 = quarters
dimes = 4
nickels = 7

Answer 7

Q = Quarters
D = Dimes
N = Nickels
Eqn #1: Q + D + N = 14
Eqn #2: D = Q + 1 or Q = D - 1
Eqn #3: N = D + 3
Sub into Eqn #1
Q + D + N = 14
(D -1 ) + D + (D + 3) = 14
3D + 2 = 14
3D = 12
D = 4 therefore Q = 3 and N = 7

Answer 8

Set up your basic equation:
n + d + q = 14

Then set up the other equations, which you will use to substitute variables out of the original equation until you are left with just one variable:
d = q + 1
n = d + 3

Get everything in terms of one variable, which in this case would be easiest to do with d:
d = q + 1
d - 1 = q

Substitute:
n + d + q = 14
(d + 3) + d + (d - 1) = 14
3d + 2 = 14
3d = 12
d = 4

Now that you've solved for d, plug that value back into your second and third equations to solve for the other two variables:
n = 4 + 3
4 = q + 1

Answer 9

d=q+1
n=d+3
n=q+4
n+d+q=14 3 equations & 3 unknowns

substitute q+1 for d & q+4 for n
q+4+q+1+q=14
3q=5=14
3q=9
q=3
d=q+1=3+1=4
n=q+4=3+4=7

Answer 10

(1+q)+(3+d)+q=14