Algebra World Problem #2

Guest
26 mins ago     

Jack and Betty have 28 coins that are nickels and dimes. If the value of the coins is $1.95. How many nickles do they have?

 

N+D=28   so   N=28-D

 

 

5N+10D=195

N+2D=39

 

28-D+2D=39

28+D=39

D=39-28

D=11

N=28-11=17

 

11 dimes and 17 nickels

Let N=number of nickels, then we have:

28 - N=number of dimes, but,

.05N + .10(28 - N)=1.95

.05 + 2.8 -.10N=1.95

-.05N=-.85 divide both sides by -.05

N=-.85/-.05

N=17 number of nickles that Jack and Betty have, and,

28-17=11 number of dimes that Jack and Betty have, because:

(17 X .05) + (11 X .10)=$1.95