Question

In the following exercises,translate to a system of equations and solve.

Priam has dimes and pennies in a cup holder in his car.The total value of coins is $4.21.The number of dimes is three less than four times the number of pennies.How many dimes and how many pennies are in the cup?

Short answer

The number of dimes are 41.And, the number of pennies are 11.

Step-by-step solution

Step 1.Given information

Let x represents the number dimes.

Let y represents the number of pennies.

Since the value of each dime is $0.10

And value of each pennie is $0.01

The total values of coins in the car is about $4.21

That means,0.10x+0.01y=4.21

The number of dimes is three less than four times the number of pennies

That means,x=4y-3.

Step 2.To find the number of dimes and number of pennies.

Consider the system of linear equations.

$$\begin{gathered} x=4y-3 \\ 0.10x+0.01y=4.21 \end{gathered}$$

the object is to solve the system of linear equations by substitution:

Step 3.Substitute x=4y-3 in the second equation 0.10x+0.01y=4.21 and solve for y.

$$\begin{gathered} 0.10x+0.01y=4.21 \\ 0.10(4y-3)+0.01y=4.21 \\ 0.40y-0.30+0.01y=4.21 \\ 0.41y=4.21+0.30 \\ 0.41y=4.51 \\ y=\frac{4.51}{0.41} \\ y=11 \end{gathered}$$

Step 4.Substitute y=11  into x=4y-3  and solve for x.

$$\begin{gathered} x=4\left(11\right)-3 \\ x=44-3 \\ x=41 \end{gathered}$$

Step 5.Check

First substitute $x=41,y=11$into the equation $x=4y-3$.

$$\begin{gathered} x=4y-3 \\ 41=4\left(11\right)-3 \\ 41=44-3 \\ 41=41 \end{gathered}$$

This is true.

Also substitute $x=41,y=11$into the equation $0.10x+0.01y=4.21$.

$$\begin{gathered} 0.10x+0.01y=4.21 \\ 0.10\left(41\right)+0.01\left(11\right)=4.21 \\ 4.1+0.11=4.21 \\ 4.21=4.21 \end{gathered}$$

This is true.

Therefore,The number of dimes are 41.And, the number of pennies are 11.