Answer:
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.