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2.4: Expressing Words as Algebraic Expressions

Difficulty Level: At Grade Created by: CK-12

Using Mathematical Symbols to Represent Words

Objectives

The lesson objectives for Expressing Words as Algebraic Expressions are:

• Using mathematical symbols to represent words
• Using algebraic expressions to represent words
• Writing linear equations in one variable to represent words

Introduction

Mathematics is a language like English or French or Chinese is a language. Mathematics can be seen as a universal language. What this means is that words spoken in mathematical terms mean the same in the United States as they mean in Canada, as they mean in Europe, and so forth. For example, if you were to say “I am going to add 3 avocados to 2 avocados, the word add is a universal term. Someone may not know what an avocado is but they would know what the term “add” means. The term “add” would give you the mental image of the symbol “+”.

In this lesson, you are going to explore some of these mathematical symbols and their word translations. Words such as gain, more, sum, total, increase, plus all mean to add. Words such as difference between, minus, decrease, less, fewer, and loss all mean to subtract. Words such as the product of, double \begin{align*}(2x)\end{align*}, twice \begin{align*}(2x)\end{align*}, triple \begin{align*}(3x)\end{align*}, a fraction of, a percent of, or times all mean to multiply. And finally words such as the quotient of, divided equally, and per mean to divide. Knowing these terms will lead to the next lesson on translating algebraic expressions from words to symbols.

Watch This

Guidance

Rob is describing his weight training to his friend James. He said that when he started training he weighed 185 pounds. He gained 8 pounds in the first month of training. How much did he weigh?

The word gain is the same as saying add.

Therefore Rob weighs \begin{align*}185 + 8 = 193 \ pounds\end{align*}.

Example A

What is the sum of five and seventeen?

Break apart the sentence. It is often helpful to underline the words before and after the word AND. Also, it is helpful to circle the mathematical symbol.

\begin{align*}& \text{What is the} \ \boxed{\text{sum}} \ \text{of} \ \underline{\text{five}} \ \text{and} \ \underline{\text{seventeen}}? \\ &\qquad \qquad \qquad \uparrow \qquad \quad \uparrow \qquad \quad \ \ \uparrow\\ & \qquad \qquad \qquad \ {\color{red}+} \qquad \ \ 5 \qquad \quad \ 17\end{align*}

Then translate the symbols together into a mathematical equation and solve it.

\begin{align*}5 {\color{red}+} 17 = 22\end{align*}

Example B

Thomas had twenty four dollars and after shopping his money decreased by four dollars.

Break apart the sentence. Underline the words before and after the word AND, and circle the mathematical symbol.

\begin{align*}& \text{Thomas had} \ \underline{\text{twenty-four}} \ \text{dollars and after shopping his money} \ \boxed{\text{decreased}} \ \text{by} \ \underline{\text{four}} \ \text{dollars.}\\ &\qquad \qquad \qquad \quad \uparrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \uparrow \qquad \qquad \ \ \uparrow\\ & \qquad \qquad \qquad \quad 24 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ {\color{red}-} \qquad \qquad \ 4\end{align*}

Then translate the symbols together into a mathematical equation and solve it.

\begin{align*}24 {\color{red}-}4 = 20\end{align*}

Therefore Thomas had 20.00 left after shopping. Example C Nick, Chris, and Jack are sharing a bag of jelly beans. There are 30 jelly beans for the three boys to share equally. How many would each get? Again, break apart the sentence. Underline the words before and after the word AND, and circle the mathematical symbol. \begin{align*}& \text{There are} \ \underline{30} \ \text{jelly beans for the} \ \underline{\text{three}} \ \text{boys to} \ \boxed{\text{share equally.}} \ \text{How many would each get?}\\ &\qquad \qquad \ \uparrow \qquad \qquad \qquad \qquad \qquad \uparrow \qquad \qquad \qquad \ \ \uparrow\\ &\qquad \qquad \ 30 \qquad \qquad \qquad \qquad \quad \ \ 3 \qquad \qquad \qquad \ \ {\color{red}\div}\end{align*} \begin{align*}30 \div 3 = 10\end{align*} Therefore each boy would get 10 jelly beans. Vocabulary Addition Words such as gain, more, sum, total, increase, plus all mean to addition or to add. Subtraction Words such as the difference between, minus, decrease, less, fewer, loss all mean subtraction or to subtract. Multiplication Words such as the product of, double \begin{align*}(2x)\end{align*}, twice \begin{align*}(2x)\end{align*}, triple \begin{align*}(3x)\end{align*}, fraction of, percent of, times all mean multiplication or to multiply. Division Words such as the quotient of, divided equally, and per mean division or to divide. Guided Practice 1. What is twelve increased by eighteen? 2. Joanne and Jillian were each going to share their babysitting money for the week. They made45.00 in total. How much does each girl receive?
3. The number five is increased by seven. Three-fourths of this number is then decreased from twenty. What is the result?

1. What is twelve increased by eighteen?

Break apart the sentence. Underline the words before and after the word AND, and circle the mathematical symbol.

\begin{align*}& \text{What is} \ \underline{\text{twelve}} \ \boxed{\text{increased}} \ \text{by} \ \underline{\text{eighteen}}?\\ &\qquad \qquad \ \uparrow \qquad \qquad \uparrow \qquad \qquad \quad \uparrow\\ &\qquad \qquad \ 12 \qquad \quad \ {\color{red}+} \qquad \qquad \quad 18\end{align*}

\begin{align*}12 + 18 = 30\end{align*}

2. Joanne and Jillian were each going to share their babysitting money for the week. They made 45.00 in total. How much does each girl receive? Break apart the sentence. Underline the words before and after the word AND, and circle the mathematical symbol. \begin{align*}& \underline{\text{Joanne and Jillian}} \ \text{were each going to} \ \boxed{\text{share}} \ \text{their babysitting money for the week. They made} \ \underline{\45.00} \ \text{in total}?\\ &\qquad \qquad \uparrow \qquad \qquad \qquad \qquad \qquad \quad \quad \ \ \uparrow \qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \qquad \qquad \qquad \ \uparrow\\ &\qquad \qquad \ 2 \qquad \qquad \qquad \qquad \qquad \quad \quad \ \ {\color{red}\div} \qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \qquad \qquad \quad \ \ 45\end{align*} \begin{align*}45 \div 2 = 22.5\end{align*} Therefore each girl would get22.50.

3. The number five is increased by seven. Three-fourths of this number is then decreased from twenty. What is the result?

Break apart the sentence. Underline the words before and after the word AND, and circle the mathematical symbol.

\begin{align*}& \text{The number} \ \underline{\text{five}} \ \text{is} \ \boxed{\text{increased}} \ \text{by} \ \underline{\text{seven}}. \ \boxed{\text{Three-fourths}} \ \text{of this} \ \underline{\text{number}} \ \text{is then} \ \boxed{\text{decreased}} \ \text{from} \ \underline{\text{twenty}}.\\ &\qquad \qquad \qquad \uparrow \qquad \quad \uparrow \qquad \qquad \quad \ \uparrow \qquad \qquad \quad \uparrow \qquad \qquad \qquad \quad \ \uparrow \qquad \qquad \qquad \quad \ \uparrow \qquad \qquad \qquad \uparrow\\ &\qquad \qquad \qquad 5 \qquad \quad {\color{red}+} \qquad \qquad \quad \ 7 \qquad \qquad \quad {\color{red}\times} \qquad \qquad \qquad \ \ 5+7 \qquad \qquad \qquad {\color{red}-} \qquad \qquad \quad \ 20\end{align*}

Step 1: \begin{align*}5 + 7 = 12\end{align*}

Step 2: \begin{align*}20- \frac{3}{4}(12) = 11\end{align*}

Summary

Knowing how to translate key words from English into mathematical symbols is important in problem solving. The first step in any problem solving situation in mathematics is always to read the problem. Afterwards, translating the words into mathematical symbols is next. In other words identifying the mathematical key words that indicate what mathematical operation you are to perform and with what numbers to use in the operation.

Within the mathematical operations, of course, there are a variety of terms that can be used for addition, subtraction, multiplication, and division. Experience and practice with problem solving will help better acquaint you with the key words that translate into these operations.

Problem Set

1. Six less than fifty-three is what number?
2. Twice the sum of eight and nine is what number?
3. Twenty-five is diminished by four times five. What is the result?
4. The product of five times four plus seven is what number?
5. The sum of forty-four and fifty-two then divided by twelve results in what number?

1. Six less than fifty-three is what number? Break apart the sentence. Underline the words before and after the word AND, and circle the mathematical symbol.

\begin{align*}& \underline{\text{Six}} \ \boxed{\text{less}} \ \text{than} \ \underline{\text{fifty-three}} \ \text{is what number}?\\ & \uparrow \quad \ \ \uparrow \qquad \qquad \quad \uparrow\\ & 6 \quad \ \ {\color{red}-} \qquad \qquad \ \ 53\end{align*}

\begin{align*}53 - 6 = 47\end{align*}

1. Twenty-five is diminished by four times five. What is the result? Break apart the sentence. Underline the words before and after the word AND, and circle the mathematical symbol.

\begin{align*}& \underline{\text{Twenty-five}} \ \text{is} \ \boxed{\text{diminished}} \ \text{by} \ \underline{\text{four}} \ \boxed{\text{times}} \ \underline{\text{five}}.\\ & \uparrow \qquad \qquad \qquad \quad \uparrow \qquad \qquad \quad \ \ \uparrow \qquad \ \uparrow \qquad \uparrow\\ & 25 \qquad \qquad \qquad \ \ {\color{red}-} \qquad \qquad \quad \ 4 \qquad \ {\color{red}\times} \qquad 5\end{align*}

Step 1: \begin{align*}4 \times 5 = 20\end{align*}

Step 2: \begin{align*}25 - 20 = 5\end{align*}

1. The sum of forty-four and fifty-two then divided by twelve results in what number? Break apart the sentence. Underline the words before and after the word AND, and circle the mathematical symbol.

\begin{align*}& \text{The} \ \boxed{\text{sum}} \ \text{of} \ \underline{\text{forty-four}} \ \text{and} \ \underline{\text{fifty-two}} \ \text{then} \ \boxed{\text{divided by}} \ \underline{\text{twelve}} \ \text{results in what number?}\\ & \qquad \quad \uparrow \qquad \qquad \uparrow \qquad \qquad \quad \uparrow \qquad \qquad \qquad \quad \uparrow \qquad \qquad \uparrow\\ & \qquad \quad {\color{red}+} \qquad \quad \ \ 44 \qquad \qquad \ \ 52 \qquad \qquad \qquad \ \ {\color{red}\div} \qquad \quad \ \ 12\\\end{align*}

Step 1: \begin{align*}44 + 52 = 96\end{align*}

Step 2: \begin{align*}96 \div 12 = 8\end{align*}

Using Algebraic Expressions to Represent Words

Introduction

The previous concept introduced you to the keywords used in translating word problems into mathematical expressions. This lesson was essential in order to move onto algebraic expressions. An algebraic expression contains not only numbers but also contains variables as you remember from previous concepts.

In this lesson, you will learn to translate algebraic expressions. Remember in the previous lessons, you solved equations with one variable. These were algebraic expressions. Algebraic expressions can also be grouped (having parentheses) and therefore need to use the distributive property. However, as with the previous lesson, you need to look for key words and translate them into mathematical expressions.

Watch This

Guidance

The sum of two consecutive even integers is 34. What are the integers?

Let \begin{align*}x =\end{align*} integer 1

Then \begin{align*}x + 2 =\end{align*} integer 2 (Because they are even, they must be separated by at least one number. For example: 2, 4, 6, 8,... all have one number in between them.)

You can know write an algebraic expression to solve for the two integers. Remember that you are talking about the sum. You can square the operation in the question.

The \begin{align*}\boxed{\text{sum}}\end{align*} of two consecutive even integers is 34.

\begin{align*}x + (x + 2) &= 34\\ x + x + 2 &= 34 && (\approx \text{Remove the brackets})\\ 2x + 2 &= 34 && (\approx \text{Combine like terms})\\ 2x + 2 {\color{red}-2} &= 34 {\color{red}-2} && (\approx \text{Subtract} \ 2 \ \text{from both sides to isolate the variable})\\ 2x &= 32 && (\approx \text{Simplify})\\ \frac{2x}{{\color{red}2}} &= \frac{32}{{\color{red}2}} && (\approx \text{Divide both sides by} \ 2 \ \text{to solve for the variable})\\ x &= 16 && (\approx \text{Simplify})\end{align*}

Therefore the first integer is 16 and the second integer is \begin{align*}(16 + 2) = 18\end{align*}. Also \begin{align*}16 + 18\end{align*} is indeed 34.

Example A

Two consecutive integers have a sum of 173. What are those numbers?

Let \begin{align*}x =\end{align*} integer 1

Then \begin{align*}x + 1 =\end{align*} integer 2 (Because they are consecutive, they must be separated by at only one number. For example: 1, 2, 3, 4,... all are consecutive.)

You can square the operation in the question.

\begin{align*}\boxed{ \ \ \ \ }\end{align*}

Two consecutive integers have a sum of 173.

\begin{align*}x + (x + 1) &= 173\\ x + x + 1 &= 173 && (\approx \text{Remove the brackets})\\ 2x + 1 &= 173 && (\approx \text{Combine like terms})\\ 2x + 1 {\color{red}-1} &= 173 {\color{red}-1} && (\approx \text{Subtract} \ 1 \ \text{from both sides to isolate the variable})\\ 2x &= 172 && (\approx \text{Simplify})\\ \frac{2x}{{\color{red}2}} &= \frac{172}{{\color{red}2}} && (\approx \text{Divide both sides by} \ 2 \ \text{to solve for the variable})\\ x &= 86 && (\approx \text{Simplify})\end{align*}

Therefore the first integer is 86 and the second integer is \begin{align*}(86 + 1) = 87\end{align*}. Also \begin{align*}86 + 87\end{align*} is indeed 173.

Example B

When a number is subtracted from 35, the result is 11. What is the number?

Let \begin{align*}x =\end{align*} number

You can square the operation in the question.

When a number is \begin{align*}\boxed{\text{subtracted}}\end{align*} from 35, the result is 11.

\begin{align*}35 - x &= 11\\ 35 {\color{red}- 35} - x &= 11 {\color{red}- 35} && (\approx \text{Subtract} \ 35 \ \text{from both sides to isolate the variable})\\ -x &= -24 && (\approx \text{Simplify})\\ \frac{-x}{{\color{red}-1}} &= \frac{-24}{{\color{red}-1}} && (\approx \text{Divide both sides by} \ -1 \ \text{to solve for the variable})\\ x &= 24 && (\approx \text{Simplify})\end{align*}

Therefore the number is 24.

Example C

When one third of a number is subtracted from one half of a number, the result is 14. What is the number?

Let \begin{align*}x =\end{align*} number

You can square the operation in the question.

When one third of a number is \begin{align*}\boxed{\text{subtracted}}\end{align*} from one half of a number, the result is 14

\begin{align*}\frac{1}{2}x-\frac{1}{3}x=14\end{align*}

You need to get a common denominator in this problem in order to solve it. For this problem, the denominators are 2, 3, and 1. The LCD is 6. Therefore multiply the first fraction by \begin{align*}\frac{3}{3}\end{align*}, the second fraction by \begin{align*}\frac{2}{2}\end{align*}, and the third number by \begin{align*}\frac{6}{6}\end{align*}.

\begin{align*}\left({\color{red}\frac{3}{3}}\right) \frac{1}{2}x-\left({\color{red}\frac{2}{2}}\right) \frac{1}{3}x &= \left({\color{red}\frac{6}{6}}\right)14\\ \frac{3}{6}x-\frac{2}{6}x &= \frac{84}{6} && (\approx \text{Simplify})\end{align*}

Now that the denominator is the same, the equation can be simplified to be:

\begin{align*}3x-2x &= 84\\ x &= 84 && (\approx \text{Combine like terms})\end{align*}

Therefore the number is 84.

Vocabulary

Algebraic Expression
An algebraic expression contains numbers, variables and operations.
Consecutive
The term consecutive means in a row. Therefore an example of consecutive numbers is 1, 2, and 3. An example of consecutive even numbers would be 2, 4, and 6. An example of consecutive odd numbers would be 1, 3, and 5.

Guided Practice

1. What is a number that when doubled would equal sixty?
2. The sum of two consecutive odd numbers is 176. What are these numbers?
3. The perimeter of a rectangular frame is 48 in. What are the lengths of each side?

1. What is a number that when doubled would equal sixty?

Let \begin{align*}x =\end{align*} number

You can square the operation in the question.

What is a number that when \begin{align*}\boxed{\text{doubled}}\end{align*} would equal sixty?

\begin{align*}2x &= 60\\ \frac{2x}{{\color{red}2}} &= \frac{60}{{\color{red}2}} && (\approx \text{Divide by} \ 2 \ \text{to solve for the variable})\\ x &= 30 && (\approx \text{Simplify})\end{align*}

Therefore the number is 30.

2. The sum of two consecutive odd numbers is 176. What are these numbers?

Let \begin{align*}x =\end{align*} first number

Let \begin{align*}x + 2 =\end{align*} second number

You can square the operation in the question.

The \begin{align*}\boxed{\text{sum}}\end{align*} of two consecutive odd numbers is 176.

\begin{align*}x + (x + 2) &= 176\\ x + x + 2 &= 176 && (\approx \text{Remove brackets})\\ 2x + 2 &= 176 && (\approx \text{Combine like terms})\\ 2x+2 {\color{red}-2} &= 176 {\color{red}-2} && (\approx \text{Subtract} \ 2 \ \text{from both sides of the equal sign to isolate the variable})\\ 2x &= 174 && (\approx \text{Simplify})\\ \frac{2x}{{\color{red}2}} &-\frac{174}{{\color{red}2}} && (\approx \text{Divide by} \ 2 \ \text{to solve for the variable})\\ x &= 87\end{align*}

Therefore the first number is 87 and the second number is \begin{align*}(87 + 2) = 89\end{align*}.

3. The perimeter of a rectangular frame is 48 in. What are the lengths of each side?

You have to remember that a square has 4 sides of equal length in order to solve this problem.

Let \begin{align*}s =\end{align*} side length

\begin{align*}s + s + s + s &= 48 && (\approx \text{Write initial equation with four sides adding to the perimeter})\\ 4s &= 48 && (\approx \text{Simplify})\\ \frac{4s}{{\color{red}4}} &= \frac{48}{{\color{red}4}} && (\approx \text{Divide by} \ 4 \ \text{to solve for the variable})\\ s &= 12\end{align*}

Therefore the side length is 12 inches.

Summary

In this lesson you have expanded on your ability to translate mathematical expressions to include variables. In other words, you have learned how to translate algebraic expressions from words into symbols. The same rules apply as was learned in the previous lesson in terms of looking for key words to indicate the operation used in the problem.

Once the equation is known, to solve the problem you use the same rules as when solving equations with one variable. Isolating the variable and then solving for it making sure that whatever you do to one side of the equal sign you do to the other side. Drawing a diagram is also helpful in solving some of the word problems involving algebraic expressions.

Problem Set

1. The sum of two consecutive numbers is 159. What are these numbers?
2. The sum of three consecutive numbers is 33. What are these numbers?
3. A new computer is on sale for 30% off. If the sale price is $500, what was the original price? 4. Jack and his three friends are sharing an apartment for the next year while at university (8 months). The rent cost$1200 per month. How much does Jack have to pay?
5. You are designing a triangular garden with an area of 168 square feet and a base length of 16 feet. What would be the height of the triangle of the garden shape?

1. The sum of two consecutive numbers is 159. What are these numbers? Let \begin{align*}x =\end{align*} first number Let \begin{align*}x + 1 =\end{align*} second number You can square the operation in the question. The \begin{align*}\boxed{\text{sum}}\end{align*} of two consecutive numbers is 159.

\begin{align*}x + (x + 1) &= 159\\ x + x + 1 &= 159 && (\approx \text{Remove brackets})\\ 2x + 1 &= 159 && (\approx \text{Combine like terms})\\ 2x + 1 {\color{red}- 1} &= 159 {\color{red}- 1} && (\approx \text{Subtract} \ 1 \ \text{from both sides of the equal sign to isolate the variable})\\ 2x &= 158 && (\approx \text{Simplify})\\ \frac{2x}{{\color{red}2}} &-\frac{158}{{\color{red}2}} && (\approx \text{Divide by} \ 2 \ \text{to solve for the variable})\\ x &= 79\end{align*}

Therefore the first number is 79 and the second number is \begin{align*}(79 + 1) = 80\end{align*}.

1. A new computer is on sale for 30% off. If the sale price is 500, what was the original price? Let \begin{align*}p =\end{align*} original price of the computer 30% off = multiply by 0.70 (since you will be still paying 70% of the regular price) Sale price =500

\begin{align*}0.70p &= \500 && (\approx \text{Put information together to write algebraic expression})\\ \frac{0.70p}{{\color{red}0.70}} &= \frac{\500}{{\color{red}0.70}} && (\approx \text{Divide by} \ 0.30 \ \text{to solve for the variable})\\ p &= 714.29\end{align*}

Therefore the regular price of the computer is 714.29. 1. You are designing a triangular garden with an area of 168 square feet and a base length of 16 feet. What would be the height of the triangle of the garden shape? Area of garden \begin{align*}= 168 \ ft^2\end{align*} Base length (b) = 16 ft \begin{align*}Area &= \frac{1}{2}(bh) && (\approx \text{Formula for area of triangle})\\ 168 &= \frac{1}{2} (16h) && (\approx \text{put in information you know})\\ \left({\color{red}\frac{2}{2}}\right)168 &= \frac{1}{2}(16h) && (\approx \text{Multiply the right side by} \ \frac{2}{2} \ \text{to get the same denominator on both sides of the equal sign.})\\ \frac{336}{2} &= \frac{1}{2}(16h) && (\approx \text{Simplify})\end{align*} Since both sides have the same denominator the equation can be simplified to be: \begin{align*}336 &= 16h\\ \frac{336}{{\color{red}16}} &= \frac{16h}{{\color{red}16}}&& (\approx \text{Divide by} \ 16 \ \text{to solve for the variable})\\ h &= 21\end{align*} Therefore the measure of the height in the garden plan will be 21 feet. Writing Linear Equations in One Variable to Represent Words Introduction A linear equation in one variable is an algebraic expression where the equation can be written in the form of \begin{align*}ax + b = c\end{align*}. As well, for linear equations in this form, \begin{align*}a \neq 0\end{align*}. As well, there is only one solution for these variables. This solution is often called the root of the linear equation. For the past series of lessons you have been solving linear equations in one variable. You have been learning rules where you maintain the balance from one side of the equal sign to the other side. In other words, if you add a number to one side of the equal sign, you must add the same number to the other side. The reason is to maintain the balance of the equal sign and isolate the variable in order to solve for it. In this lesson, you will be performing these same type of problem solving strategies but first you will have to use your skills of translating them from English into math. Watch This Guidance One forth of a number less one sixth of three less this same number is equal to one. What is the number? If first let’s write down what you know: Let \begin{align*}x =\end{align*} the number \begin{align*}& \underline{\text{One forth}} \ \text{of a number} \ \boxed{\text{less}} \ \underline{\text{one sixth}} \ \text{of} \ \underline{\text{three less this same number}} \ \text{is equal to} \ \underline{\text{one}}.\\ & \quad \ \ \uparrow \qquad \qquad \qquad \qquad \ \ \uparrow \quad \ \ \quad \uparrow \qquad \qquad \qquad \qquad \quad \uparrow \qquad \qquad \qquad \qquad \qquad \quad \ \ \uparrow\\ & \quad \quad \frac{1}{4} \qquad \qquad \qquad \qquad \ {\color{red}-} \quad \ \quad \frac{1}{4} \qquad \qquad \qquad \ \ \quad (x-3) \qquad \qquad \qquad \qquad \quad \quad \ 1\end{align*} The equation would therefore be: \begin{align*}\frac{1}{4}x-\frac{1}{6}(x-3)=1\end{align*} The common denominator for 4, 6, and 1 is 12. Therefore multiply the first fraction by \begin{align*}\frac{3}{3}\end{align*}, the second fraction by \begin{align*}\frac{2}{2}\end{align*}, and the third number by \begin{align*}\frac{12}{12}\end{align*}. \begin{align*}\left({\color{red}\frac{3}{3}}\right) \frac{1}{4}x-\left({\color{red}\frac{2}{2}}\right) \frac{1}{6}(x-3) &= \left({\color{red}\frac{12}{12}}\right)1\\ \frac{3}{12}x -\frac{2}{12} (x-3) &= \frac{12}{12}\\ 3x-2(x-3) &=12 && (\approx \text{Simplify since the denominators are the same})\\ 3x-2x+6 &= 12 && (\approx \text{Remove brackets})\\ x+6 &= 12 && (\approx \text{Combine like terms})\\ x+6 {\color{red}-6} &= 12 {\color{red}-6} && (\approx \text{Subtract} \ 6 \ \text{from both sides to solve for the variable})\\ x &= 6\end{align*} Therefore \begin{align*}x = 6\end{align*}. Example A If four times a number is added to six, the result is 50. What is that number? Let \begin{align*}n =\end{align*} the number \begin{align*}& \text{If} \ \underline{\text{four}} \ \boxed{\text{times}} \ \text{a number is} \ \boxed{\text{added}} \ \text{to} \ \underline{\text{six}}, \ \text{the result is} \ 50.\\ & \quad \ \uparrow \qquad \uparrow \qquad \qquad \qquad \qquad \ \ \uparrow \quad \quad \quad \uparrow\\ & \quad \ \ 4 \qquad {\color{red}\times} \qquad \qquad \qquad \qquad {\color{red}+} \quad \quad \quad 6\end{align*} The equation would therefore be: \begin{align*}4x+6 &= 50\\ 4x+6 {\color{red}-6} &= 50 {\color{red}-6} && (\approx \text{Subtract} \ 6 \ \text{from both sides to isolate the variable})\\ 4x &= 44 && (\approx \text{Simplify})\\ \frac{4x}{{\color{red}4}} &= \frac{44}{{\color{red}4}} && (\approx \text{Divide by} \ 4 \ \text{to solve for the variable})\\ x &= 11\end{align*} Therefore \begin{align*}x = 11\end{align*}. Example B This week, Emma earned ten more than half the number of dollars she earned last week babysitting. If this week, she earned 100 dollars, how much did she earn last week? Let \begin{align*}d =\end{align*} the number of dollars earned \begin{align*}& \text{This week, Emma earned} \ \underline{\text{ten}} \ \boxed{\text{more}} \ \text{than half the number of dollars she earned last week babysitting.}\\ & \qquad \qquad \qquad \qquad \qquad \quad \ \ \uparrow \quad \ \ \uparrow \qquad \qquad \ \ \uparrow\\ & \qquad \qquad \qquad \qquad \qquad \quad \ \ 10 \quad \ {\color{red}+} \qquad \qquad \ \frac{1}{2}\end{align*} The equation would therefore be: \begin{align*}\frac{1}{2}d+10 &= 100\\ \frac{1}{2}d+10 {\color{red}-10} &= 100 {\color{red}-10} && (\approx \text{Subtract} \ 16 \ \text{from both sides to isolate the variable})\\ \frac{1}{2}d &= 90 && (\approx \text{Simplify})\\ ({\color{red}2}) \frac{1x}{2} &= 90 ({\color{red}2}) && (\approx \text{Multiply by} \ 2 \ \text{to solve for the variable})\\ x &= 180\end{align*} Therefore last week, Emma earned180 babysitting.

Example C

Three is twenty-one divided by the sum of a number plus five.

Let \begin{align*}t =\end{align*} the number

\begin{align*}& \ \underline{\text{Three}} \ \text{is} \ \underline{\text{twenty-one}} \ \boxed{\text{divided}} \ \text{by the} \ \boxed{\text{sum of}} \ \text{a number plus} \ \underline{\text{five}}.\\ & \quad \uparrow \qquad \quad \ \quad \uparrow \qquad \qquad \ \uparrow \qquad \qquad \qquad \ \ \uparrow \qquad \qquad \uparrow \qquad \qquad \ \uparrow\\ & \quad \ 3 \qquad \quad \quad 21 \qquad \qquad {\color{red}\div} \qquad \qquad \qquad \ {\color{red}+} \qquad \qquad t \qquad \qquad \ 5\end{align*}

The equation would therefore be:

\begin{align*}3=\frac{21}{(t+5)}\end{align*}

Since our unknown is on the denominator, let’s multiply both sides by \begin{align*}(t+5)\end{align*} to eliminate the denominator from the left side.

\begin{align*}{\color{red}(t+5)}3 &= \frac{21}{\cancel{(t+5)}} \cancel{{\color{red}(t+5)}}\\ (t+5)3 &= 21 && (\approx \text{Simplify})\\ 3t+15 &= 21 && (\approx \text{Remove brackets})\\ 3t+15 {\color{red}-15} &= 21 {\color{red}-15} && (\approx \text{Subtract} \ 15 \ \text{from both sides to isolate the variable})\\ 3t &= 6 && (\approx \text{Simplify})\\ \frac{3t}{{\color{red}3}} &= \frac{6}{{\color{red}3}} && (\approx \text{Divide both sides by} \ 3 \ \text{to solve for the variable})\\ t &= 2\end{align*}

Therefore the number is 2.

Vocabulary

Linear Equation
A linear equation in one variable is an algebraic expression where the equation can be written in the form of \begin{align*}ax + b = c\end{align*}. The variable is represented by “\begin{align*}x\end{align*}” in this equation but can be represented by any letter but must be in the first power (no exponent). Therefore \begin{align*}x\end{align*} is not \begin{align*}x^2\end{align*}, or \begin{align*}x^3\end{align*}. Because the \begin{align*}x\end{align*} is in the first power, it makes the algebraic expression with the variable linear. As well, for linear equations in this form, \begin{align*}a \neq 0\end{align*}. As well, there is only one solution for these variables.
Root
A root in a linear equation is the solution. For example in the linear equation \begin{align*}p + 3 = 5\end{align*}, the root (or the solution) would be 2.

Guided Practice

1. Five less than three times a number is forty-six.
2. Hannah had $237 in her bank account at the start of the summer. She worked for four weeks and now she has$1685 in the bank. How much did Hannah make each week in her summer job?
3. The formula to estimate the length of the Earth's day in the future is found to be twenty–four hours added to the number of million years divided by two hundred and fifty. In five hundred million years, how long will the Earth's day be?

1. Five less than three times a number is forty six.

Let \begin{align*}f =\end{align*} the number

\begin{align*}& \text{\underline{Five}} \ \boxed{\text{less}} \ \text{than} \ \text{\underline{three}} \ \boxed{\text{times}} \ \text{a} \ \text{\underline{number}} \ \text{is} \ \text{\underline{forty-six}}.\\ & \ \uparrow \qquad \uparrow \qquad \qquad \ \uparrow \qquad \ \uparrow \qquad \quad \ \ \uparrow \qquad \qquad \ \uparrow\\ & \ \ 5 \quad \ \ - \qquad \qquad 3 \qquad {\color{red}\times} \qquad \quad \ \ f \qquad \qquad 46\end{align*}

The equation would therefore be:

\begin{align*}3f-5 &= 46\\ 3f-5 {\color{red}+5} &= 46 {\color{red}+5} && (\approx \text{Add} \ 5 \ \text{to both sides to isolate the variable})\\ 3f &= 51 && (\approx \text{Simplify})\\ \frac{3f}{{\color{red}3}} &= \frac{51}{{\color{red}3}} && (\approx \text{Divide both sides by} \ 3 \ \text{to solve for the variable})\\ f &= 17\end{align*}

Therefore the number is 17.

2. Hannah had $237 in her bank account at the start of the summer. She worked for four weeks and now she has$1685 in the bank. How much did Hannah make each week in her summer job?

Let \begin{align*}h =\end{align*} the number of dollars earned in one week

Starts with 237 Works for 4 weeks making \begin{align*}h\end{align*} dollars Ends with1685

The equation would therefore be:

\begin{align*}4h+237 &= 1685\\ 4h+237 {\color{red}-237} &= 1685 {\color{red}-237} && (\approx \text{Subtract} \ 237 \ \text{from both sides to isolate the variable})\\ 4h &= 1448 && (\approx \text{Simplify})\\ \frac{4h}{{\color{red}4}} &= \frac{1448}{{\color{red}4}} &&(\approx \text{Divide by} \ 4 \ \text{to solve for the variable})\\ h&=362\end{align*}

Therefore each week, Hannah earns 362. 3. The formula to estimate the length of the Earth's day in the future is found to be twenty–four hours added to the number of million years divided by two hundred and fifty. In five hundred million years, how long will the Earth's day be? Let \begin{align*}t =\end{align*} the number years Length of day \begin{align*}= d\end{align*} \begin{align*}& \text{The formula to estimate the length of the Earth's day in the future is found to be} \ \underline{\text{twenty-four}}\\ & \text{hours} \ \boxed{\text{added}} \ \text{to the number of years} \ \boxed{\text{divided}} \ \text{by} \ \underline{\text{two hundred and fifty}}.\\ & \qquad \qquad \uparrow \qquad \qquad \qquad \uparrow \qquad \qquad \quad \quad \quad \uparrow \qquad \qquad \qquad \qquad \uparrow \qquad \qquad \qquad \qquad \qquad \quad \ \uparrow\\ & \qquad \qquad {\color{red}+} \qquad \qquad \quad \quad t \qquad \qquad \quad \quad \quad {\color{red}\div} \qquad \qquad \qquad \quad \ 250 \qquad \qquad \qquad \ \ \qquad \qquad 24\\ & \text{In} \ \underline{\text{five hundred million}} \ \text{years, how long will the Earth's day be?}\\ & \qquad \qquad \quad \uparrow\\ & \qquad \qquad \quad \ t\end{align*} The equation would therefore be: \begin{align*}d &= 24+\frac{1}{250}(500)\\ d &= 24+2 && (\approx \text{Multiply through to solve for the variable})\\ d &= 26 && (\approx \text{Simplify})\end{align*} Therefore in 500 million years, the length of the day will be 26 hours. Summary In this final lesson on representing words into linear equations, you have been able to use the skills of selecting key words and translating them into mathematical operations, variables, and numbers in order to solve for the variables. All of the concepts you have learned in the previous nine lessons have been helpful, in one or another, in the translation and solving of these algebraic expressions. It is important for you to look for the key words, translate them into a linear equation, and then solve the algebraic equation by isolating the variable then solving for it. Problem Set 1. Three times a number less six is one hundred twenty-six. 2. Sixty dollars was two-thirds the total money spent by Jack and Thomas at the store. 3. Ethan mowed lawns for five weekends over the summer. He worked ten hours each weekend and each lawn takes an average of two and one-half hours. How many lawns did Ethan mow? 4. The area of a rectangular pool is found to be two hundred eighty square feet. If the base length of the pool is 20 feet, what is the width of the pool? 5. A cell phone company charges a base rate of10 per month plus 5¢ per minute for any long distance calls. Sandra gets her cell phone bill for 21.20. How many long distance minutes did she use? Answers 1. Three times a number less six is one hundred twenty-six. Let \begin{align*}j =\end{align*} the number \begin{align*}& \underline{\text{Three}} \ \boxed{\text{times}} \ \text{a number} \ \boxed{\text{less}} \ \underline{\text{six}} \ \text{is} \ \underline{\text{one hundred twenty-six}}.\\ & \quad \uparrow \qquad \ \uparrow \qquad \quad \ \uparrow \quad \qquad \uparrow \quad \ \uparrow \qquad \qquad \qquad \ \ \uparrow\\ & \quad \ 3 \qquad {\color{red}\times} \quad \ \qquad j \qquad \ \ - \quad 6 \qquad \qquad \qquad 136\end{align*} The equation would therefore be: \begin{align*}3j-6 &= 126\\ 3j-6 {\color{red}+6} &= 126 {\color{red}+6} && (\approx \text{Add} \ 6 \ \text{to both sides to isolate the variable})\\ 3j &= 132 && (\approx \text{Simplify})\\ \frac{3j}{{\color{red}3}} &=\frac{132}{{\color{red}3}} && (\approx \text{Divide both sides by} \ 3 \ \text{to solve for the variable})\\ j &= 44\end{align*} Therefore the number is 44. 1. Ethan mowed lawns for five weekends over the summer. He worked ten hours each weekend and each lawn takes an average of two and one-half hours. How many lawns did Ethan mow? Let \begin{align*}m =\end{align*} number of lawns mowed Mowed lawns for five weekends Worked 10 hours each weekend Each lawn takes 2.5 hours The equation would therefore be: \begin{align*}m &= \frac{(5 \times 10)}{2.5} && (\approx \text{Simplify to solve for the variable})\\ m &= \frac{(50)}{2.5}\\ m &= 20\end{align*} Therefore in the five weekends, Ethan would be able to mow 20 lawns. 1. A cell phone company charges a base rate of10 per month plus 5¢ per minute for any long distance calls. Sandra gets her cell phone bill for 21.20. How many long distance minutes did she use? Let \begin{align*}m =\end{align*} number of long distance minutes Total bill =21.20

A cell phone company charges a base rate of \$10 per month plus 5¢ per minute for any long distance calls.

The equation would therefore be:

\begin{align*}21.20 &= 10+0.05m\\ 21.20 {\color{red}-10} &= 10 {\color{red}-10}+0.05m && (\approx \text{Subtract} \ 10 \ \text{from both sides to isolate the variable})\\ 11.20 &= 0.05m && (\approx \text{Simplify})\\ \frac{11.20}{{\color{red}0.05}} &= \frac{0.05m}{{\color{red}0.05}} && (\approx \text{Divide by} \ 0.05 \ \text{to solve for the variable})\\ m &= 224\end{align*}

Therefore Sandra talked for 224 long distance minutes.

Summary

In this lesson, you have learned how to translate from words into mathematical symbols in order to solve linear equations. Remember the key steps are to identify key words in the problem and from here write your linear equation. Sometimes it is helpful to circle mathematical operations and underline constants and variables to help you in the identification of key words.

Just as there are different definitions for words in the English language, there are different words that can be used for the mathematical operations. You will be more able to identify these with experience in solving problems.

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