Suppose that Maria has 140 coins consisting of pennies, nickels, and dimes. The number of nickels she has is 9 less than twice the number of pennies; the number of dimes she has is 19 less than three times the number of pennies. How many coins of each kind does she have?

p = pennies

2p - 9 = nickles
3p - 19 = dimes

p + 2p - 9 + 3p - 19 = 140
6p - 28 = 140
6p = 168
p = 28

p = pennies = 28
2p - 9 = nickles = 47
3p - 19 = 65

there are 100 coins. ten are dimes. the rest are pennies and nickles. there are twice as many pennies as nickles. how many pennies and nickles are in the jar?

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?

To solve this problem, we can use a system of equations. Let's define the variables:

Let P represent the number of pennies.
Let N represent the number of nickels.
Let D represent the number of dimes.

We know that Maria has a total of 140 coins, so we can write the first equation:

P + N + D = 140

We also have two additional pieces of information:

1. "The number of nickels she has is 9 less than twice the number of pennies."
This can be written as: N = 2P - 9

2. "The number of dimes she has is 19 less than three times the number of pennies."
This can be written as: D = 3P - 19

Now we can solve this system of equations.

Substitute the expressions for N and D into the first equation:

P + (2P - 9) + (3P - 19) = 140

Combine like terms:

6P - 28 = 140

Add 28 to both sides:

6P = 168

Divide both sides by 6:

P = 28

Now substitute the value of P back into the expressions for N and D:

N = 2(28) - 9 = 47
D = 3(28) - 19 = 65

Therefore, Maria has 28 pennies, 47 nickels, and 65 dimes.