Q. 363 - Intermediate Algebra

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?

Step-by-Step Solution

Answer

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

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 = 4 y - 3 in the second equation 0 . 10 x + 0 . 01 y = 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 = 4 y - 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.