Victor has quarters, dimes and pennies in his pocket. He has three times fewer pennies than dimes and he has 4 fewer quarters than pennies. How many coins does he have total if they are worth $4.60?

Let's define the variables :

Q = number of quarters that Victor has

P = number of pennies that victor has

D = number of dimes that victor has.

We know that:

"He has three times fewer pennies than dimes"

P = D*3

"he has 4 fewer quarters than pennies."

Q = P - 4

And we know that in total, he has $4.60, then:

Q*$0.25 + P*$0.10 + D*0.05 = $4.60

Then we have a system of equations:

P = D/3

Q = P - 4

Q*$0.25 + P*$0.01 + D*0.10 = $4.60

To solve this, we need to replace variables in order to eliminate them.

We can start by replacing the first equation into the other two:

Q = D/3 - 4

Q*$0.25 + (D/3)*$0.01 + D*0.10 = $4.60

Now we can replace the top equation into the bottom one:

(D/3 - 4)*$0.25 + (D/3)*$0.01 + D*0.10 = $4.60

And now solve this for D:

D*($0.25/3 + $0.01/3 + $0.10) -$1  = $4.60

D*$0.1866... = $4.60 + $1.00 = $5.60

D = $5.60/$0.1866... = 30

Then he has 30 dimes.

P = D/3 = 30/3 = 10 pennies

Q = P - 4 = 10 - 4 = 6 quarters.

In total he has 30 + 10 + 6 = 46 coins.

because 1.2 isnt half of 6 duh

1 and 0

step-by-step explanation: