Adam has $2.15 in nickels, dimes and quarters in his pocket. if the number of dimes is 4 less than the number of nickels, and 5 more than the number of quarters, then how many of each type of coin does he have?

8 Dimes,12 Nickels, and 3 Quarters

Step-by-step explanation:

d= amount of dimes  n=amount of nickels  q=amount of quarters

You have to make formulas for the different variables. Since there are four more dimes than nickels so you have d+4=n. same thing with quarters but LESS so you would subtract d-5=q.

Then making the big picture;

Dimes have 10 cents Nickles have 5 cents, and Quarters have 25 cents. We do not know how many of each we have but we do know how much they are so when you add them together you would have to get the amount of money. That is  were the amount of each coin comes in so we have 10(d) for dimes 5(n) for nickles then 25(q) for quarters. We know the total of those is going to get 215 for the $2.15. This is how we get the big formula 10(d)+5(n)+25(q)=215.

Now we just plug things in 10(d)+5(d+4)+25(d-5)=215. Now we solve for D.

Distribute the 5 and 25   10d+5d+20+25d-125=215

Add common multiples   40d-105=215

Add 105 to both sides   40d=320

Divide 40 to both sides d=8

Now you have to figure out how many quarters and nickels you have just plug them into the other formulas;

n=d+4    n=8+4    n=12

q=d-5     q=8-5     q=3

You can ALWAYS check you answer by plug them into the "big picture"

10d+5n+25q=215   ~   10(8)+5(12)+25(3)=215  ~  80+60+75=215

probably the last option

step-by-step explanation: