Nickles, dimes and quarters add up to $2.35. Dimes is three less than sum of nickels and quarters. How many of each type is there if there's 19 total?

Question

Set up and solve a system of 3 equations for the following word problems.

Answer 1

Hi Kimberly, to start, let d be the dimes, n the nickels, q the quarters.

 

Equation 1:

 

The sum of the coins is equal to 19.

 

n + d + q = 19

 

Equation 2:

 

The value of the coins is $2.35.

 

.05n + .10d + .25q = 2.35

 

Equation 3:

 

The dimes is three less than the sun of the nickels and quarters.

 

d = n + q - 3

 

Solve:

 

    n  +     d +      q =    19

.05n + .10d + .25q = 2.35

   -n +      d -       q =    -3

 

Multiply the second equation by 100 to clear decimals.

 

     n +     d +    q =   19
   5n + 10d + 25q = 235
   -n +     d -      q =   -3

 

Substitute for n + q - 3 for d in the first and second equations.

 

    n + n + q - 3 + q = 19

 

    2n  + 2q = 22

 

Divide by 2:

 

     n  + q  = 11

 

In the second equation:

 

     5n + 10(n + q - 3) + 25q = 235

 

     5n + 10n + 10q - 30 + 25q = 235

 

    15n + 35q = 265

 

 We now have two equations in two unknowns:

 

       n +    q =   11

   15n + 35q = 265

 

Multiply the first equation by -15 and add:

 

    -15n  - 15q  = -165

     15n  +35q  =  265

   -----------------------

               20q  = 100

 

Divide both sides by 20:

 

q = 5

 

Substitute for q in n + q = 11

 

n + 5 = 11

 

Subtract 5 from both sides:

 

n = 6

 

Return to n + d + q = 19, to solve for d:

 

6 + d + 5 = 19

 

d = 8

 

Solution:

 

nickels = 6, dimes = 8, and quarters = 5

 

check:

 

n + d + q = 19

 

6 + 8 + 5 = 19

 

19 = 19, values check.

 

.05n + .10d + .25q = 2.35

 

.05 * 6 + .10 * 8 + .25 * 5 = 2.35

 

.30 + .80 + 1.25 = 2.35

 

2.35 = 2.35, values check.

 

d = n + q - 3

 

8 = 6 + 5 - 3

 

8 = 8, values check.

 

Questions?

 

 

 

 

Answer 2

If we represent the number or nickles with "n", dimes with "d", and quarters with "q," we get the following equation because each nickel is worth $0.05, dimes are worth $0.10 and quarters worth $0.25:

.05n+.1d+.25q = 2.35

 

The second sentence tells us n + q - 3 (nickels and quarters minus 3) = d

 

n+q-3 = d

 

and the final sentence gives us

 

n+q+d = 19

So our three equations are

 

.05n+.1d+.25q = 2.35

n+q-3 = d

n+q+d = 19

 

if we substitute the second equation into the third, we get the equation:

n+q+ (n+q-3) = 19

 

since n+q-3 = d

 

Rewriting this by combining like terms we get

2n + 2q = 22

 

Simplify

 

n + q = 11

 

since we know n+q - 3 = d, we can use this to get that d = 8

 

knowing that n = 11-q and d = 8, we plug these into the first equation

.05(11-q) + .01(8) + .25(q) = 2.35

 

distribute

 

.55-.05q + .80 + .25q = 2.35

 

combine like terms

 

1.35 + .20q = 2.35

 

.20q = 1

 

q = 5

 

know we know q=5, d=8, plugging these into any equation gives us that n = 6