Q. 133 - Intermediate Algebra

Question

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

Sherri saves nickels and dimes in a coin purse for her daughter. The total value of the coins in the purse is $$0.95$. The number of nickels is two less than five times the number of dimes. How many nickels and how many dimes are in the coin purse?

Step-by-Step Solution

Answer

In the coin purse, there are $13$ nickels and $3$ dimes.

Step 1. Given Information

It is given that the total value of the nickels and dimes in the purse is $$0.95$.

Also, the number of nickels is two less than five times the number of dimes. 

Step 2. Identify and name what we are looking for

We need to find the number of nickels and dimes in the coin purse.

Let $x$ represents the number of nickels and $y$ represents the number of dimes.

Step 3. Form the equations

The value of one nickel is $$0.05$ and the value of one dime is $$0.10$. The value of $x$ nickels and $y$ dimes is $$0.95$, so an equation can be written as

$0.05x+0.10y=0.95 ...\left(1\right)$

The number of nickels is two less than five times the number of dimes, so

$x=5y-2 ...\left(2\right)$

Step 4. Solve using substitution

Using the second equation, substitute $5y-2$ for $x$ in the first equation and solve for $y$

$\begin{array}{rcl}0.05x+0.10y& =& 0.95\\ 0.05(5y-2)+0.10y& =& 0.95\\ 0.25y-0.10+0.10y& =& 0.95\\ 0.25y+0.10y-0.10+0.10& =& 0.95+0.10\\ 0.35y& =& 1.05\\ \frac{0.35y}{0.35}& =& \frac{1.05}{0.35}\\ y& =& 3\end{array}$

Step 5. Solve for x

Substitute $3$ for $y$ in the second equation and solve for $x$

$$\begin{gathered} x=5y-2 \\ x=5\times 3-2 \\ x=15-2 \\ x=13 \end{gathered}$$

So the number of nickels is $13$ and the number of dimes is $3$.

Step 6. Check the solution

Substitute $13$ for $x$ and $3$ for $y$ in the first equation formed.

$\begin{array}{rcl}0.05x+0.10y& =& 0.95\\ 0.05·13+0.10·3& =& 0.95\\ 0.65+0.30& =& 0.95\\ 0.95& =& 0.05\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}x& =& 5y-2\\ 13& =& 5·3-2\\ 13& =& 15-2\\ 13& =& 13\end{array}$

This is also a true statement.

So the point satisfies both the equations.