1/3 as many dimes as nickels
3 more quarters than nickels
17 coins in all
Q=___ D=___ N=___


Sagot :

What you have to do in this problem is create an algebraic equation with one variable. Because this set of problems show relationship between the other dimes and quarters with nickels, X in your equation should represent nickels. Your equation should look something like this: X=number of nickels X+3=number of quarters 1/3x=number of dimes Equation: X+X+3+1/3X=17 7/3X+3=17 7/3X=14 X=6 From here you can plug the number of nickels back into the original equations to find that there are 9 quarters, 2 dimes, and 6 nickels.