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

Answer:

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

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

Step-by-step 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) :

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

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

$q(0.2)=2.6$

$q=\frac{2.6}{0.2}=13$

$q=13$

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 .