Chuin has 18 coins in his pocket. He has nickels and dimes that total $1.45. How many nickels and how many dimes does he have? Write and solve a system of equations to find the answer.

A. 10 Nickles, 8 dimes
B. 9 nickles, 9 dimes
C. 8 nickles, 10 dimes
D. 7 nickles, 11 dimes

Respuesta :

Answer:

D

Step-by-step explanation:

Let n represent the number of nickels and d represent the number of dimes Chuin has.

Since Chuin has a total of 18 coins in his pocket, the sum of the number of nickels and dimes must total 18. So:

[tex]n+d=18[/tex]

The total worth of the coins is $1.45. Since each nickel is worth $0.05 and each dime is worth $0.10:

[tex]0.05n+0.1d=1.45[/tex]

Solve the system. We can use elimination (you can use substitution if you prefer, it works just as perfect!). First, multiply both sides of the second equation by 10:

[tex]0.5n+d=14.5[/tex]

Next, we can multiply the first equation by negative one:

[tex]-n-d=-18[/tex]

Add the two equations together:

[tex](0.5n+d)+(-n-d)=(14.5)+(-18)[/tex]

Simplify:

[tex]-0.5n=-3.5[/tex]

Divide both sides by -0.5. So:

[tex]n=7[/tex]

Chuin has seven nickels.

Using the first equation again, substitute n for 7 and solve for d:

[tex](7)+d=18\Rightarrow d=11[/tex]

Therefore, Chuin has seven nickels and 11 dimes.

Our answer is D.

D, (7*5=35)+(11*10=110)=145