# 3.11: Solve Multi-Step Equations Involving Rational Numbers

**At Grade**Created by: CK-12

**Practice**Equations with Decimals, Fractions, and Parentheses

Jose has been playing the French horn for many years and until now, everything has been easy. Now Mrs. Kline, the band director has assigned him a new piece of music to work on and it is very tricky. Jose has been practicing the new piece.

His best rehearsal was on Saturday, when he practiced for 90 minutes. On Sunday, he had a birthday party to go to, so he did not practice as long. On Monday, he had a math test to study for and so he practiced for half as long as he did on Monday.

When Jose went to band practice on Tuesday afternoon, he struggled through the piece.

“How long did you practice?” Mrs. Kline asked him.

“Well, from Saturday to Tuesday I practiced a total of 3 hours,” Jose said.

**If this is true, how long did Jose practice on Monday and Tuesday? You will need to write an equation and solve it to figure out the answer to this problem. Jose needs to practice his French horn a bit more and you will need to use the information taught in this Concept to help you figure out each dilemma.**

### Guidance

Rational numbers include integers, fractions, and terminating decimals. Some equations may require you to work with a combination of these kinds of numbers. If you know how to solve an equation, you can apply the same rules when you work with rational numbers.

Take a look at this dilemma.

*Solve for \begin{align*}b\end{align*} b: \begin{align*}-6 \left( 1 - \frac{b}{12} \right) = \frac{2}{3}\end{align*}−6(1−b12)=23*

**This problem involves two different kinds of rational numbers: integers (-6 and 1) and fractions \begin{align*}\left ( \frac{b}{12} \right .\end{align*} (b12 and \begin{align*}\left . \frac{2}{3} \right )\end{align*}23). You will need to know how to compute with fractions as well as how to compute with integers in order to solve this.**

**Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by -6 and then subtract those products.**

\begin{align*}-6 \left (1 - \frac{b}{12} \right ) &= \frac{2}{3}\\
(-6 \times 1) - \left ( -6 \times \frac{b}{12} \right ) &= \frac{2}{3}\\
-6 - \left ( \frac{-6}{1} \times \frac{1}{12}b \right ) &= \frac{2}{3}\\
-6 - \left ( \frac{-6}{12}b \right ) &= \frac{2}{3}\\
-6 - \left ( - \frac{6}{12}b \right ) &= \frac{2}{3}\\
-6 + \left ( \frac{6}{12}b \right ) &= \frac{2}{3}\end{align*}

You may recognize immediately that the variable term, \begin{align*}\frac{6}{12}b\end{align*}

\begin{align*}-6 + \frac{6}{12}b &= \frac{2}{3}\\
-6 + \frac{1}{2}b &= \frac{2}{3}\end{align*}

**Now, we can solve as we would solve any two-step equation. To get \begin{align*}\frac{1}{2}b\end{align*} 12b by itself on one side of the equation, we can subtract -6 from both sides.**

\begin{align*}-6 + \frac{1}{2}b &= \frac{2}{3}\\
-6 - (-6) + \frac{1}{2}b &= \frac{2}{3} - (-6)\\
-6 + 6 + \frac{1}{2}b &= \frac{2}{3} + 6\\
0 + \frac{1}{2}b &= 6 \frac{2}{3}\\
\frac{1}{2}b &= 6 \frac{2}{3}\end{align*}

To get \begin{align*}b\end{align*}

\begin{align*}\frac{1}{2}b &= 6 \frac{2}{3}\\ \frac{1}{2}b \times \frac{2}{1} &= 6 \frac{2}{3} \times \frac{2}{1}\\ \frac{\bcancel{1}}{\bcancel{2}}b \times \frac{\bcancel{2}}{\bcancel{1}} &= \frac{20}{3} \times \frac{2}{1}\\ 1b &= \frac{40}{3}\\ b &= 13 \frac{1}{3}\end{align*}

**The value of \begin{align*}b\end{align*} is \begin{align*}13 \frac{1}{3}\end{align*}.**

**Now, let's solve an algebraic equation that includes both decimals and fractions.**

*Solve for \begin{align*}k\end{align*}: \begin{align*}0.4k + 0.2k + \frac{3}{10} = \frac{9}{10}\end{align*}.*

**First, add the like terms \begin{align*}0.4k\end{align*} and \begin{align*}0.2k\end{align*} on the left side of the equation.**

\begin{align*}0.4k + 0.2k + \frac{3}{10} &= \frac{9}{10}\\ 0.6k + \frac{3}{10} &= \frac{9}{10}\end{align*}

**The next step is to isolate the term with the variable, \begin{align*}0.6k\end{align*}, on one side of the equation. We can do this by subtracting \begin{align*}\frac{3}{10}\end{align*} from both sides of the equation.**

\begin{align*}0.6k + \frac{3}{10} &= \frac{9}{10}\\ 0.6k + \frac{3}{10} - \frac{3}{10} &= \frac{9}{10} - \frac{3}{10}\\ 0.6k + 0 &= \frac{6}{10}\\ 0.6k &= \frac{6}{10}\end{align*}

**Since \begin{align*}0.6k\end{align*} means \begin{align*}0.6 \times k\end{align*}, we should divide each side of the equation by 0.6 to get the \begin{align*}k\end{align*} by itself on one side of the equation.** This will involve dividing a fraction, \begin{align*}\frac{6}{10}\end{align*}, by a decimal, 0.6. To do this, you will need to convert both numbers to the same form. One way to do this would be to convert the fraction \begin{align*}\frac{6}{10}\end{align*} to a decimal. \begin{align*}\frac{6}{10}\end{align*} is read as “six tenths,” so the decimal form of \begin{align*}\frac{6}{10}\end{align*} is 0.6.

\begin{align*}0.6k &= \frac{6}{10}\\ 0.6k &= 0.6\\ \frac{0.6k}{0.6} &= \frac{0.6}{0.6}\\ 1k &= 1\\ k &= 1\end{align*}

**The value of \begin{align*}k\end{align*} is 1.**

#### Example A

\begin{align*}8.7n - 3.2n + 4.5 = 37.5\end{align*}

**Solution: \begin{align*}n =6\end{align*}**

#### Example B

\begin{align*}\frac{x}{.9} = -72\end{align*}

**Solution: \begin{align*}x=-64.8\end{align*}**

#### Example C

\begin{align*}17x - 22.3x + 4 = -33.1\end{align*}

**Solution: \begin{align*}x = 7\end{align*}**

Now let's go back to the dilemma at the beginning of the Concept.

**First, write an equation to show what you know and what you don’t know.**

**Saturday = 90 minutes**

**\begin{align*}\text{Monday} = t - \text{missing time}\end{align*}**

**\begin{align*}\text{Tuesday} = \frac{1}{2} t - \text{half the time of Monday}\end{align*}**

**Total time = 3 hours**

\begin{align*}90 + t + \frac{1}{2} t = 3 \ hours\end{align*}

**First, convert hours to minutes.**

\begin{align*}90 + t + \frac{1}{2} t = 180 \ minutes\end{align*}

**Now we can solve the equation.**

**Monday’s time = 60 minutes**

**Tuesday’s time = 30 minutes**

### Vocabulary

- Integer
- the set of whole numbers and their opposites.

- Rational Numbers
- a set of numbers that includes integers, decimals, fractions, terminating and repeating decimals. These numbers can be written in fraction form.

- Fraction
- a part of a whole written using a numerator and a denominator.

- Decimal
- a part of a whole written using place value and a decimal point.

- Repeating Decimal
- a decimal where the digits repeat in a pattern and eventually end.

- Terminating Decimal
- a decimal where the digits eventually end, but where numbers do not repeat in a pattern.

### Guided Practice

Here is one for you to try on your own.

For a long-distance call, Guillermo's phone company charges $0.10 for the first minute and $0.05 for each minute after that. Guillermo was charged $1.00 for a long distance call he made last Friday.

a. Write an algebraic equation that could be used to represent \begin{align*}m\end{align*}, the length in minutes of Guillermo's $1.00 long-distance call.

b. Determine how many minutes his $1.00 long-distance call lasted.

**Solution**

Consider part *a* first.

You know that the phone company charges $0.10 for the first minute and $0.05 for each minute after that. How could you represent that? If the company charged $0.05 for each minute the call lasted, you could represent that as \begin{align*}0.05 \times m\end{align*}. However, the company charges $0.10 for the first minute and $0.05 for each minute *after* that first minute.

So, a 1-minute call will cost: \begin{align*}\$0.10 + (\$0.05 \times 0) = \$0.10 + \$0.00 = \$0.10\end{align*}.

A 2-minute call will cost: \begin{align*}\$0.10 + (\$0.05 \times 1) = \$0.10 + \$0.05 = \$0.15\end{align*}.

A 3-minute call will cost: \begin{align*}\$0.10 + (\$0.05 \times 2) = \$0.10 + \$0.10 = \$0.20\end{align*}.

Notice that the number you multiply by $0.05 is always 1 less than the length of the call, in minutes. If \begin{align*}m\end{align*} represents the length of a call in minutes, then this could be represented as: \begin{align*}\$0.10 + \$0.05 \times (m - 1)\end{align*}.

**Write an equation that could be used to represent the cost of Guillermo's $1.00 call.**

\begin{align*}& (\text{cost of first minute}) + (\text{cost of each minute after first minute}) = (\text{total cost})\\ & \qquad \qquad \downarrow \ \ \qquad \qquad \downarrow \qquad \qquad \qquad \qquad \downarrow \qquad \quad \quad \quad \qquad \qquad \ \downarrow \qquad \ \downarrow\\ & \qquad \qquad 0.10 \qquad \ \quad + \qquad \qquad \qquad \quad 0.05(m-1) \quad \qquad \qquad \ = \quad 1.00\end{align*}

So, the equation \begin{align*}0.10 + 0.05(m-1) = 1.00\end{align*} represents the number of minutes that Guillermo's $1.00 phone call lasted.

**Next, consider part** *b***.**

To find the length of the $1.00 call in minutes, solve the equation for \begin{align*}m\end{align*}. First, apply the distributive property to the right side of the equation.

\begin{align*}0.10 + 0.05(m-1) &= 1.00\\ 0.10 + (0.05 \times m) - (0.05 \times 1) &= 1.00\\ 0.10 + 0.05m - 0.05 &= 1.00\end{align*}

Use the commutative property to rearrange the terms being added so it is easier to see how to combine the like terms. Then combine the like terms.

\begin{align*}(0.10 + 0.05m) - 0.05 &= 1.00\\ (0.05m + 0.10) - 0.05 &= 1.00\\ 0.05m + (0.10 - 0.05) &= 1.00\\ 0.05m + 0.05 &= 1.00\end{align*}

**Now, solve as you would solve any two-step equation.** First, subtract 0.05 from both sides of the equation.

\begin{align*}0.05m + 0.05 &= 1.00\\ 0.05m + 0.05 - 0.05 &= 1.00 - 0.05\\ 0.05m + 0 &= 0.95\\ 0.05m &= 0.95\end{align*}

Next, divide both sides of the equation by 0.05.

\begin{align*}0.05m &= 0.95\\ \frac{0.05m}{0.05} &= \frac{0.95}{0.05}\\ 1m &= 19\\ m &= 19\end{align*}

**The value of \begin{align*}m\end{align*} is 19, so the $1.00 call lasted for 19 minutes.**

### Video Review

Khan Academy Solving Linear Equations 4

### Practice

Directions: Solve each equation to find the value of the variable.

- \begin{align*}7n - 3.2n + 6.5 = 17.9\end{align*}
- \begin{align*}0.2(3 + p) = -5.6\end{align*}
- \begin{align*}s + \frac{3}{5} + \frac{1}{5} = 1\frac{2}{5}\end{align*}
- \begin{align*}j + \frac{5}{7} - \frac{1}{7} = 9\frac{4}{7}\end{align*}
- \begin{align*}\frac{3}{4} \left( g - \frac{1}{2} \right ) = \frac{1}{8}\end{align*}
- \begin{align*}-2 \left ( 1 - \frac{a}{4} \right ) = \frac{1}{8}\end{align*}
- \begin{align*}0.09y - 0.08y = .005\end{align*}
- \begin{align*}.36x + 2.55x = -8.55\end{align*}
- \begin{align*}\frac{1}{3}y + \frac{1}{3}y = 8\end{align*}
- \begin{align*}\frac{1}{4}x + \frac{1}{3} = \frac{2}{3}\end{align*}
- \begin{align*}\frac{1}{2}x = 18\end{align*}
- \begin{align*}.9x = 56\end{align*}
- \begin{align*}.6x + 1 = 19\end{align*}
- \begin{align*}\frac{1}{4}x + 2 = 19\end{align*}
- \begin{align*}9.05x = 27.15\end{align*}

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Decimal

In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).distributive property

The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, .fraction

A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a*rational number*.

Integer

The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...rational number

A rational number is a number that can be expressed as the quotient of two integers, with the denominator not equal to zero.Repeating Decimal

A repeating decimal is a decimal number that ends with a group of digits that repeat indefinitely. 1.666... and 0.9898... are examples of repeating decimals.Terminating Decimal

A terminating decimal is a decimal number that ends. The decimal number 0.25 is an example of a terminating decimal.### Image Attributions

Here you'll learn to solve multi-step equations involving rational numbers.