SOLUTION: Malik has $3.08 in dimes, pennies,and nickels. He has nine fewer nickels than dimes and one more peeny than dimes. How manof each coin does he have.

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Malik has $3.08 in dimes, pennies,and nickels. He has nine fewer nickels than dimes and one more peeny than dimes. How manof each coin does he have.      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 15367: Malik has $3.08 in dimes, pennies,and nickels. He has nine fewer nickels than dimes and one more peeny than dimes. How manof each coin does he have.
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Since nickels are given in terms of dimes and pennies are in terms of dimes, let x = the number of dimes.

Let x = number of dimes
x-9 = number of nickels
x+1 = number of pennies

Equation is the VALUE of the coins, which is the number of coins times the value of each one, and express it in cents, which would be 308 cents.

10x + 5(x-9) + 1(x+1) = 308
10x + 5x - 45 + x + 1 = 308
16x - 44 = 308

Add 44 to each side:
16x - 44 + 44 = 308 + 44
16x = 352

Divide both sides by 16
x= 352/16 = 22 Dimes
x-9 = 22-9 = 13 Nickels
x+1 = 23 Pennies

Check: Find the total value of the coins.
22 Dimes + 13 Nickels + 23 Pennies
$2.20 + .65 + .23 = $3.08
It checks!

R^2 at SCC