Suppose that Maria has 140 coins consisting of pennies, nickels, and dimes. The number of nickels she has is 12 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?

p+n+d = 140

n = 2p-12
d = 3p-22

(p,n,d) = (29,46,65)

Thank you !

To find out how many coins of each kind Maria has, we can set up a system of equations based on the given information.

Let's denote the number of pennies as 'P', the number of nickels as 'N', and the number of dimes as 'D'.

According to the problem:
1) The total number of coins is 140:
P + N + D = 140

2) The number of nickels is 12 less than twice the number of pennies:
N = 2P - 12

3) The number of dimes is 22 less than three times the number of pennies:
D = 3P - 22

Now, we have a system of three equations with three unknowns. We can solve this system to find the values of P, N, and D.

Let's solve it step by step:

From equation (2), we can express N in terms of P:
N = 2P - 12

Next, substitute this expression for N in equation (1):
P + (2P - 12) + D = 140

Combine like terms:
3P + D = 152

Now, substitute the expression for D from equation (3) into the above equation:
3P + (3P - 22) = 152

Combine like terms again:
6P - 22 = 152

Add 22 to both sides of the equation:
6P = 174

Divide both sides by 6:
P = 29

Now that we have the value of P, we can substitute it back into the other equations to find the values of N and D.

From equation (2):
N = 2P - 12
N = 2(29) - 12
N = 58 - 12
N = 46

From equation (3):
D = 3P - 22
D = 3(29) - 22
D = 87 - 22
D = 65

So, Maria has 29 pennies, 46 nickels, and 65 dimes.