Chad has a coin collection. He has 21 coins in all. He has 4 times as many pennies as nickels. He has 1 more dime than the number of nickels. He has 1 less quarter than the number of nickels. How many of each coin does he have?

12 pennies

4 dimes

3 nickels

2 quarters

Step-by-step explanation:

Let p = pennies

d = dimes

q = quarters

n - nickels

d + n + p + q = 21

p = 4n

d = n + 1

q = n - 1

Now just subsitute them in and add the terms

(n + 1) + n + (4n) + (n - 1) = 21

7n = 21

Divide both sides by 7

7n/7 = 21/7

n = 3

Now use the new n to plug in the rest of the equations

p = 4(3) = 12

d = 3 + 1 = 4

q = 3 - 1 = 2

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,