33 coins, consisting of nickels, dimes, and quarters, have a total value of $3.30. If there are three times as many nickels as quarters, and one-half as many dimes as nickels. how many coins are there

Answer 1

DNE, real solution does not exist

{ }, null set

because the 4 equations are inconsistent/contradictory

no solution satisfies all 4 equations, unless one or more of the equations is linearly dependent (a multiple) of another equation.

n=3q

q= n/3

d/2=n

d = 2n

n+d+q = 33, eventually replace d and q by their equivalents in n

330 = 5n +10d +25q

divide by 5

66 = n+2d +5q here's the replacement stage

66 = n +2(n) + n/3

multiply by 3

3(66) = 3n +6n +n = 10n

n = 3(66)/10 = 198/10 =19.8 nickels

d = 2n = 39.6 dimes

q= n/3 = 6.6 quarters

n+d+q = 67 = more than 33

seemingly likely mistake(s) above somewhere

or maybe the problem is the problem using fractional parts of coins

and 4 equations which can't all 4 be satisfied by 3 variable values

odds are the problem has a typo,misprint or was mis copied

try matrix algebra, it's impossible to get a diagonal of 1's with 0's elsewhere and the solution in the augmented matrix' last column

n d q

1 0 0 0

1 -1.5 0 0

1 1 1 33

5 10 25 330

or

graph the 4 equations.

graph the two equations n+d+q = 33 and 5n +10d + 25q, intersection point is (3, 29, 1) but it doesn't satisfy the other two equations d/2 = n and n=3q