Sam found a bunch of change under his bed. He has a pile of quarters, a pile of dimes and a pile of nickels. He has the same number of quarters, dimes and nickels. When he adds it all up, he has eight dollars and eighty cents. How many of each coin does Sam have?

In this concept, you will learn to solve multi-step equations involving decimals.

### Multi Step Equations with Decimals

**Integers** include positive whole numbers (1, 2, 3, 4, 5, . . .), their opposites (-1, -2, -3, -4, -5, . . .), and zero. Integers are **rational numbers**. A **rational number** is any number that can be written as the ratio of two integers or you can think of this in fraction form. So, an integer such as -3, which can be written as the ratio \begin{align*}\frac{-3}{1}\end{align*}, is a rational number.

A **fraction**, such as , can be written as the ratio of two integers. So, fractions are rational numbers.

A **terminating decimal**, such as 0.1, is also rational because it can be written as the ratio \begin{align*}\frac{1}{10}\end{align*}.

A **repeating decimal**, such as \begin{align*}0.33\bar{3}\end{align*}, is rational because even though the digit 3 repeats over and over in the decimal form. It can be expressed as the ratio of two integers or \begin{align*}\frac{1}{3}\end{align*}. All integers, fractions, terminating decimals and repeating decimals are rational numbers. You can solve equations with other rational numbers in them.

Let’s start by looking at solving equations involving decimals.

Solve for ‘

’:\begin{align*}3x-2.5x+0.5=4.5\end{align*}

First, subtract the like terms on the left side of the equation.

\begin{align*}\begin{array}{rcl} 3x-2.5x+0.5 &=& 4.5 \\ 3.0x-2.5x+0.5 &=& 4.5\\ 0.5x+0.5 &=& 4.5 \end{array}\end{align*}

Next, isolate the term with the variable,

, on one side of the equation. Since 0.5 is added to , we should subtract 0.5 from both sides of the equation.\begin{align*}\begin{array}{rcl} 0.5x+0.5 &=& 4.5 \\ 0.5x+0.5-0.5 &=& 4.5-0.5\\ 0.5x &=& 4.0 \end{array}\end{align*}

Then, divide by 0.5 to solve for ‘\begin{align*}x\end{align*}’.

\begin{align*}\begin{array}{rcl} & \ 0.5x &=& 4.0 \\ & \frac{0.5x}{0.5} &=& \frac{4.0}{0.5}\\ & \ \ x &=& 8 \end{array}\end{align*}

The answer is 8.

### Examples

#### Example 1

Earlier, you were given a problem about Sam who found the change under his bed. He looked at the pile of nickels, dime, and quarters and noticed that he had the same number of each coin. When he added it up, he had a total of $8.80. He wants to know how many of each coin he has.

First, let ‘

’ represent the number of each coin Sam has. Remember that a quarter is 25¢ or $0.25, a dime is 10¢ or $0.10, and a nickel is 5¢ or $0.05. So the equation for Sam to solve would be:

Next, combine like terms.

Then, divide both sides by 0.4.

The answer is 22.

Sam has 22 of each type of coin.

#### Example 2

Solve for ‘\begin{align*}x\end{align*}’:\begin{align*}0.1(z-4.2)=0.48\end{align*}

First, you can see that we have parentheses in this equation. Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by 0.1.

\begin{align*}\begin{array}{rcl} 0.1(z-4.2) &=& 0.48 \\ (0.1 \times z) -(0.1 \times 4.2) &=& 0.48 \\ 0.1z-0.42 &=& 0.48 \end{array}\end{align*}

Next, solve as you would solve any two-step equation. To get

by itself on one side of the equation, add 0.42 to both sides.\begin{align*}\begin{array}{rcl} 0.1z-0.42 &=& 0.48 \\ 0.1z-0.42 +0.42 &=& 0.48 +0.42 \\ 0.1z &=& 0.90 \end{array}\end{align*}

Then, to get

by itself on one side of the equation, divide both sides by 0.1.

The answer is 9.

#### Example 3

First, to get

by itself on one side of the equation, divide both sides by 0.7.\begin{align*}\begin{array}{rcl} & 0.7x &=& 4.90 \\ & \frac{0.7x}{0.7} &=& \frac{4.90}{0.7} \\ & x &=& 7 \end{array}\end{align*}

The answer is 7.

#### Example 4

First, solve as you would solve any two-step equation. To get \begin{align*}0.3x\end{align*} by itself on one side of the equation, subtract 10 from both sides.

\begin{align*}\begin{array}{rcl} 0.3x+10 &=& 31 \\ 0.3x+10 -10 &=& 31 -10\\ 0.3x &=& 21 \end{array}\end{align*}

Next, to get ‘\begin{align*}x\end{align*}’ by itself on one side of the equation, divide both sides by 0.3.

\begin{align*}\begin{array}{rcl} &0.3x &=& \ 21 \\ &\frac{0.3x}{0.3} &=& \frac{21}{0.3} \\ &x &=& \ 70 \end{array}\end{align*}

The answer is 70.

#### Example 5

First, combine like terms on the left side of the equation.

Next, to get \begin{align*}0.38x\end{align*} by itself on one side of the equation, subtract 4 from both sides.

Then, to get ‘\begin{align*}x\end{align*}’ by itself on one side of the equation, divide both sides by 0.38.

The answer is 2.

### Review

Solve each equation to find the value of the variable.

- \begin{align*}3.2n+6.5n=38.8\end{align*}
- \begin{align*}2(a+4)+0.5a=23\end{align*}
- \begin{align*}0.54y+0.16y+0.22y=3.68\end{align*}
- \begin{align*}0.26x+0.18x=-3.08\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 3.9.