In a jar there are 73 coins consisting of quarters, dimes, and pennies. There is $9.19. The number of quarters is 2 more than twice the number of dimes. How many quarters, nickels, and dimes are there?

30 quarters, 14 dimes and 29 pennies

Step-by-step explanation:

Note: there are no nickels, just pennies. With quarters, dimes and nickels there is no solution since all three coins are multiples of 5 but the sum ($9.19) is not.

Coins:

Quarters | q = 25 cents Dimes  | d = 10 cents Pennies  | p = 1 cent Total coins: 73 Total amount:  $9.19 = 919 c

Equations:

q + d + p= 73 q = 2d + 2 25q + 10d + p = 919

Substitute q in the first equation:

2d+2 +d +p = 73  ⇒ 3d + p = 71  ⇒ p = 71 -3d

Substitute p and q in the third equation:

25q + 10d + p = 919 25(2d+2) + 10d + 71 - 3d = 919 50d + 50 + 7d = 848 57d = 798 d= 798/57d= 14

Then, finding p and q:

q = 2d + 2 = 2*14 + 2 = 30p = 71 - 3d = 71 - 3*14 = 29

So there are 30 quarters, 14 dimes and 29 pennies

Proof:

30+14+29 = 7373 = 7330*25 + 14*10 + 29 = 919919= 919

First subtract 10 from both side
y^2+6y-16=0
(y+8)(y-2)=0
-8 and 2