A coin jar contains nickels, dimes and quarters. There are 46 coins in all. There are 11 more nickels than quarters. The value of the dimes is $1.00 less than the value of the quarters. How many coins of each type are in the jar?

Answer 1

Solve the system:
n+d+q=46
n=q+11
10d=25q-100
From the 3rd -->d=2.5q-10
The 1st: (q+11)+(2.5q-10)+q=46

q=10
n=21
d=15

Answer 2

solving using the elimination method and identify the system as consistent, inconsistent or dependent.
3x-4y=8
6x-2y=10

Answer 3

Suppose that Maria has 120 coins consisting of pennies, nickels, and dimes. The number of nickels she has is 14 less than twice the number of pennies; the number of dimes she has is 22 less than three times the number of pennies. How many coins of each kind does she have?

Answer 4

p+n+d=120

n=2p-14

d=3p-22

p+(2p-14)+(3p-22)=120

6p-36=120
6p=156
p=26

n=2(26)-14
n=52-14
n=38

d=3(26)-22
d=78-22
d=56

Answer 5

P(1, 2); Q(8, 26)

Answer 6

5+6