Help with algebra world problem! Please HW due tommorow

Question

John has $2.55 in nickels and dimes. If John has 9 more dimes than nickels, how many of each coin does John have?

Answer 1

hey Max

let x be the number of nickles and y the number of dimes

we know that both nickles and dimes sum up to 2.55

so 0.05x+0.10y=2.55

We are also given that John have 9 more dimes than nickles

so y=x+9

so we the equations 0.05x+0.10y=2.55

                                 y=x+9

we use substitution to solve this system

replace y by x+9 in the first equation

0.05x+0.10(x+9)=2.55

0.05x+0.1x+0.9=2.55

0.15x=1.65

x=11

we use any of the two equation to find y

y=11+9=20

so 11 nickles and 20 dimes

11x0.05+20x0.10=0.55+2=2.55 checks √

Answer 2

N = number of nickels at $0.05 each

 

D = number of dimes at $0.10 each

 

Then

 

2.55 = 0.05*N + 0.1*D

D = N + 9

 

Substitute D in first equation and get

 

2.55 = 0.05*N + 0.1*(N + 9)

 

N = 11

 

D = N + 9 = 20