Ajar holding 36 coins, only nickels and quarters, is worth $6.20. There are eight more quarters than there are nickels. How many nickels are in the jar?

  14

Step-by-step explanation:

Let n represent the number of nickels. Then the number of quarters is 36-n. The difference is ...

  (36-n) -n = 8 . . . . . 8 more quarters than nickels

  18 -n = 4 . . . . . . . divide by 2

  14 = n . . . . . . . . add n-4 to both sides

There are 14 nickels in the jar.

_____

Additional comment

The problem supplies more information than is needed for a solution. You can work this as a "sum and difference" problem, as we have above, or you can work it as a "mixture" problem where the total coin value comes into play. Any two of the three given relations will give a solution.

  n + q = 36 . . . . . . . number of coins

  5n +25q = 620 . . . value of coins (in cents)

  q -n = 8 . . . . . . . . difference in number of coins

y= -

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this