9) A coin bank contains $3.55 in quarters, dimes, and nickels. The number of
dimes is four less than the number of quarters and the number of nickels is
seven more than the number of quarters. Find the number of quarters.

9 quarters

Step-by-step explanation:

Total cost of money in the coin bank = $3.55

Let the number of dimes be = d

Let the number of quarters be = q

Let the number of nickels be = n

Where:

1 quarter = $0.25

1 dime = $0.1

1 nickel = $0.05

Hence,

0.1(d) + 0.25(q) +0.05(n) = $3.55

In the question, we are told that:

The number of dimes is four less than the number of quarters, mathematically

d = q - 4

The number of nickels is seven more than the number of quarters, mathematically,

n = q + 7

From the question, we are ask to find q

0.1(d) + 0.25(q) +0.05(n) = $3.55

since d = q - 4

n = q + 7

We substitute

0.1(q - 4) +0.25( q )+ 0.05(q + 7) = $3.55

0.1(q - 4) +0.25( q )+ 0.05(q + 7) =$3.55

0.1q - 0.4 +0.25q + 0.05q + 0.35 = $3.55

Collect like terms

0.1q + 0.25q + 0.05q = 3.55 + 0.4 - 0.35

0.4q = 3.6

q = 3.6/0.4

q = 9

Therefore the number of quarters is 9.

add -60 to 120 which will result in : 2x+3y=180

then subtract 2x: 3y=-2x+180

lastly, divide by 2: y= -2/3x+60

hope this is the answer you're looking for: )