Respuesta :
Let's let n= the number of nickels in his pocket, and d=the number of dimes in his pocket. Nickels are worth .05 and dimes are worth .10.
Now, we can write two equations using these variables and solve them as a system of equations:
n+d=19 [the number of nickels and dimes equals 19]
.05n+.10d=1.55 [the values of nickels and dimes times their quantities equals $1.55]
100(.05n+.10d=1.55) [multiply second equation by 100 to eliminate decimals]
5n+10d=155
5(n+d=19) [multiply first equation by 5 so the 'n' terms have the same coefficient]
5n+5d=95
Now subtract the equations to eliminate n and simplify to solve for d:
5n+10d=155
(-)
5n+5d=95
5d=60
d=12. There are 12 dimes
Now substitute into the simpler equation to find number of nickels:
n+(12)=19
n=7. There are 7 nickels.
Check your work:
12 (.10) + 7(.05) = 1.20 + .35 = $1.55
Now, we can write two equations using these variables and solve them as a system of equations:
n+d=19 [the number of nickels and dimes equals 19]
.05n+.10d=1.55 [the values of nickels and dimes times their quantities equals $1.55]
100(.05n+.10d=1.55) [multiply second equation by 100 to eliminate decimals]
5n+10d=155
5(n+d=19) [multiply first equation by 5 so the 'n' terms have the same coefficient]
5n+5d=95
Now subtract the equations to eliminate n and simplify to solve for d:
5n+10d=155
(-)
5n+5d=95
5d=60
d=12. There are 12 dimes
Now substitute into the simpler equation to find number of nickels:
n+(12)=19
n=7. There are 7 nickels.
Check your work:
12 (.10) + 7(.05) = 1.20 + .35 = $1.55
