Question

Yolanda paid for her movie ticket using 28 coins, all nickels and quarters. The ticket cost $4. Which system of linear equations can be used to find the numberof nickels, n, and the number of quarters, q, Yolanda used?

Answer 1

If nickels and quarters add up to 28 coins then it would be
n + q = 28

If nickels are .05 and quarters are .25 and the total is $4 then it would be
.05n + .25q = 4

Answer 2

`\left \{ {{n+q=28} \atop {n(0.05)+q(0.25)=4}} \right.`

And the pair
`(n,q)=(15,13)` is the solution.

Explanation:

We know that Yolanda paid for her movie ticket using 28 coins. She only paid with nickels and quarters.

We also know that the ticket cost $4.

We need to form a linear equation system that solves this problem.

We have the following variables :

n : number of nickels

q : number of quarters

We know that she used 28 coins ⇒ the number of nickels plus the number of quarters must be equal to 28.

We have our first equation :

`n+q=28` (I)

For the second equation we need to use the ticket price information.

We know that the ticket cost $4 and she only paid with nickels and quarters.

Therefore we can write the following equation that relates the variable ''n'' and the variable ''q'' :

`n(0.05)+q(0.25)=4` (II)

This equation represents that the number of nickels ''n'' per its value plus the number of quarters ''q'' per its value is equal to $4 that it is the value of the movie ticket.

With (I) and (II) we form the linear equation system :

`\left \{ {{n+q=28} \atop {n(0.05)+q(0.25)=4}} \right.`

This linear equation system can be used to find the value of ''n'' and ''q''.

For example, in equation (I)

`n+q=28`

we can solve it in terms of ''n'' :

`n=28-q` (III)

If we use (III) in (II) :

`(28-q)(0.05)+q(0.25)=4`

`1.4-(0.05)q+(0.25)q=4`

`q(0.2)=2.6`

`q=(2.6)/(0.2)=13`

`q=13`

Now replacing this value of q in (III) :

`n=28-13=15`

`n=15`

We find that Yoland used 15 nickels and 13 quarters to paid the movie ticket .