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# 1.3: Patterns and Equations

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Write an equation.
• Use a verbal model to write an equation.
• Solve problems using equations.

## Introduction

In mathematics, and especially in algebra, we look for patterns in the numbers that we see. The tools of algebra assist us in describing these patterns with words and with equations (formulas or functions). An equation is a mathematical recipe that gives the value of one variable in terms of the other.

For example, if a theme park charges $12 admission, then the number of people who enter the park every day and the amount of money taken by the ticket office are related mathematically. We can write a rule to find the amount of money taken by the ticket office. In words, we might say “The money taken in dollars is (equals) twelve times the number of people who enter the park.” We could also make a table. The following table relates the number of people who visit the park and the total money taken by the ticket office. Number of visitorsMoney taken($)112224336448560672784\begin{align*}& \text{Number of visitors} & & 1 & & 2 & & 3 & & 4 & & 5 & & 6 & & 7 \\ & \text{Money taken} (\) & & 12 & & 24 & & 36 & & 48 & & 60 & & 72 & & 84 \end{align*}

Clearly, we will need a big table if we are going to be able to cope with a busy day in the middle of a school vacation!

A third way we might relate the two quantities (visitors and money) is with a graph. If we plot the money taken on the vertical axis and the number of visitors on the horizontal axis, then we would have a graph that looks like the one shown as follows. Note that this graph shows a smooth line for non-whole number values of x\begin{align*} x \end{align*} (e.g., x=2.5\begin{align*} x=2.5\end{align*}). But, in real life this would not be possible because you cannot have half a person enter the park. This is an issue of domain and range, something we will talk about in the following text.

The method we will examine in detail in this lesson is closer to the first way we chose to describe the relationship. In words we said that “The money taken in dollars is twelve times the number of people who enter the park.” In mathematical terms we can describe this sort of relationship with variables. A variable is a letter used to represent an unknown quantity. We can see the beginning of a mathematical formula in the words.

The money taken in dollars is twelve times the number of people who enter the park.

This can be translated to:

the money taken in dollars =12 ×\begin{align*}= 12 \ \times \end{align*} (the number of people who enter the park)

To make the quantities more visible they have been placed in parentheses. We can now see which quantities can be assigned to letters. First we must state which letters (or variables) relate to which quantities. We call this defining the variables:

Let x=\begin{align*}x =\end{align*} the number of people who enter the theme park.

Let y=\begin{align*}y =\end{align*} the total amount of money taken at the ticket office.

We can now show the fourth way to describe the relationship, with our algebraic equation.

y=12x\begin{align*} y=12x \end{align*}

Writing a mathematical equation using variables is very convenient. You can perform all of the operations necessary to solve this problem without having to write out the known and unknown quantities in long hand over and over again. At the end of the problem, we just need to remember which quantities x\begin{align*} x \end{align*} and y\begin{align*} y \end{align*} represent.

## Write an Equation

An equation is a term used to describe a collection of numbers and variables related through mathematical operators. An algebraic equation will contain letters that relate to real quantities or to numbers that represent values for real quantities. If, for example, we wanted to use the algebraic equation in the example above to find the money taken for a certain number of visitors, we would substitute that value in for x\begin{align*} x \end{align*} and then solve the resulting equation for y\begin{align*} y \end{align*}.

Example 1

A theme park charges 12 entry to visitors. Find the money taken if 1296 people visit the park. Let’s break the solution to this problem down into a number of steps. This will help us solve all the problems in this lesson. Step 1 Extract the important information. (money taken in dollars)(number of visitors)=12×(number of visitors)=1296\begin{align*}(\text{money taken in dollars}) & = 12 \times (\text{number of visitors})\\ (\text{number of visitors}) & = 1296\end{align*} Step 2 Translate into a mathematical equation. We do this by defining variables and by substituting in known values. Let yy=(money taken in dollars)=12×1296THIS IS OUR EQUATION.\begin{align*}\text{Let} \ y & = (\text{money taken in dollars})\\ y & = 12 \times 1296 & & \text{THIS IS OUR EQUATION}.\end{align*} Step 3 Solve the equation. y=15552Answer: The money taken is15552\begin{align*} y=15552 & & \text{Answer: The money taken is} \ \15552 \end{align*}

Step 4 Check the result.

If $15552 is taken at the ticket office and tickets are$12, then we can divide the total amount of money collected by the price per individual ticket.

(number of people)=1555212=1296\begin{align*}\text{(number of people)} = \frac{15552} {12} = 1296\end{align*}

Our answer equals the number of people who entered the park. Therefore, the answer checks out.

Example 2

The following table shows the relationship between two quantities. First, write an equation that describes the relationship. Then, find out the value of b\begin{align*} b \end{align*} when a\begin{align*} a \end{align*} is 750.

a:b:0201040206030804010050120\begin{align*}& a: & & 0 & & 10 & & 20 & & 30 & & 40 & & 50 \\ & b: & & 20 & & 40 & & 60 & & 80 & & 100 & & 120 \end{align*}

Step 1 Extract the important information. We can see from the table that every time a\begin{align*} a \end{align*} increases by 10, b\begin{align*} b \end{align*} increases by 20. However, b\begin{align*}b\end{align*} is not simply twice the value of a\begin{align*} a \end{align*}. We can see that when a=0,b=20\begin{align*} a=0, b=20\end{align*} so this gives a clue as to what rule the pattern follows. Hopefully you should see that the rule linking a\begin{align*} a \end{align*} and b.\begin{align*} b. \end{align*}

“To find a\begin{align*}a\end{align*}, double the value of a\begin{align*}a\end{align*} and add 20.”

Step 2 Translate into a mathematical equation:

Text Translates to Mathematical Expression
“To find b\begin{align*} b \end{align*} \begin{align*} \rightarrow\end{align*} b=\begin{align*} b= \end{align*}
“double the value of a\begin{align*} a \end{align*} \begin{align*} \rightarrow\end{align*} 2a\begin{align*} 2a \end{align*}
“add 20” \begin{align*} \rightarrow\end{align*} +20

b=2a+20THIS IS OUR EQUATION.\begin{align*} b=2a+20 & & \text{THIS IS OUR EQUATION}.\end{align*}

Step 3 Solve the equation.

Go back to the original problem. We substitute the values we have for our known variable and rewrite the equation.

when a is 750b=2(750)+20\begin{align*}“\text{when} \ a \ \text{is} \ 750” && \rightarrow && b=2(750)+20 \end{align*}

Follow the order of operations to solve

bb=2(750)+20=1500+20=1520\begin{align*} b& =2(750)+20\\ b& =1500+20=1520\end{align*}

Step 4 Check the result.

In some cases you can check the result by plugging it back into the original equation. Other times you must simply double-check your math. Double-checking is always advisable. In this case, we can plug our answer for b\begin{align*} b \end{align*} into the equation, along with the value for a\begin{align*} a \end{align*} and see what comes out. 1520=2(750)+20\begin{align*}1520=2(750)+20\end{align*} is TRUE because both sides of the equation are equal and balance. A true statement means that the answer checks out.

## Use a Verbal Model to Write an Equation

In the last example we developed a rule, written in words, as a way to develop an algebraic equation. We will develop this further in the next few examples.

Example 3

The following table shows the values of two related quantities. Write an equation that describes the relationship mathematically.

x\begin{align*}x-\end{align*}value y\begin{align*}y-\end{align*}value
2\begin{align*}-2\end{align*} 10
0 0
2 -10
4 -20
6 -30

Step 1 Extract the important information.

We can see from the table that y\begin{align*} y \end{align*} is five times bigger than x\begin{align*} x \end{align*}. The value for y\begin{align*} y \end{align*} is negative when x\begin{align*} x \end{align*} is positive, and it is positive when x\begin{align*} x \end{align*} is negative. Here is the rule that links x\begin{align*} x \end{align*} and y\begin{align*} y \end{align*}.

\begin{align*} y \end{align*} is the negative of five times the value of \begin{align*} x \end{align*}

Step 2 Translate this statement into a mathematical equation.

Text Translates to Mathematical Expression
\begin{align*}y\end{align*} is” \begin{align*} \rightarrow\end{align*} \begin{align*} y=\end{align*}
“negative 5 times the value of \begin{align*} x \end{align*} \begin{align*} \rightarrow\end{align*} \begin{align*} -5x \end{align*}

\begin{align*} y=-5x && \text{THIS IS OUR EQUATION}.\end{align*}

Step 3 There is nothing in this problem to solve for. We can move to Step 4.

Step 4 Check the result.

In this case, the way we would check our answer is to use the equation to generate our own \begin{align*} xy \end{align*} pairs. If they match the values in the table, then we know our equation is correct. We will substitute \begin{align*} x \end{align*} values of -2, 0, 2, 4, 6 in and solve for \begin{align*} y \end{align*}.

\begin{align*}&x=-2: && y=-5(-2) && y=+10\\ &x=0:&& y=-5(0)&&y=0\\ &x=2:&& y=-5(2)&&y=-10\\ &x=4:&& y=-5(4)&&y=-20\\ &x=6:&& y=-5(6)&&y=-30\end{align*}

Each \begin{align*} xy \end{align*} pair above exactly matches the corresponding row in the table.

Example 4

Zarina has a $100 gift card, and she has been spending money on the card in small regular amounts. She checks the balance on the card weekly, and records the balance in the following table. Week Number Balance ($)
1 100
2 78
3 56
4 34

Write an equation for the money remaining on the card in any given week.

Step 1 Extract the important information.

We can see from the table that Zarina spends 22 every week. • As the week number increases by 1, the balance decreases by 22. • The other information is given by any point (any week, balance pair). Let’s take week 1: • When (week number) = 1, (balance) = 100 Step 2 Translate into a mathematical equation. Define variables: Let \begin{align*}\text{week number} = n\end{align*} Let \begin{align*}\text{Balance} = b\end{align*} Text Translates to Mathematical Expression As \begin{align*}n\end{align*} increases by 1, \begin{align*}b\end{align*} decreases by 22 \begin{align*}\rightarrow\end{align*} \begin{align*}b=-22n+? \end{align*} The ? indicates that we need another term. Without another term the balance would be -22, -44, -66,... We know that the balance in week 1 is 100. Let's substitute that value. \begin{align*}100 = -22( 1 ) +?\end{align*} The ? number that gives 100 when 22 is subtracted from it is 122. equation is therefore: \begin{align*} b=-22n+122 & & \text{THIS IS OUR EQUATION}.\end{align*} Step 3 All we were asked to find was the expression. We weren't asked to solve it, so we can move to Step 4. Step 4 Check the result. To check that this equation is correct, we see if it really reproduces the data in the table. To do that we plug in values for \begin{align*} n \end{align*} \begin{align*}& n=1 & & \rightarrow & & b=-22(1)+122 & & \rightarrow & & b=122-22=100 \\ & n=2 & & \rightarrow & & b=-22(2)+122 & & \rightarrow & & b=122-44=78 \\ & n=3 & & \rightarrow & & b=-22(3)+122 & & \rightarrow & & b=122-66=56 \\ & n=4 & & \rightarrow & & b=-22(4)+122 & & \rightarrow & & b=122-88=34 \end{align*} The equation perfectly reproduces the data in the table. The answer checks out. Note: Zarina will run out of money on her gift card (i.e. her balance will be 0) between weeks 5 and 6. ## Solve Problems Using Equations Let’s solve the following real-world problem by using the given information to write a mathematical equation that can be solved for a solution. Example 5 A group of students are in a room. After 25 students leave, it is found that \begin{align*} \frac {2}{3} \end{align*} of the original group is left in the room. How many students were in the room at the start? Step 1 Extract the important information We know that 25 students leave the room. We know that \begin{align*} \frac {2}{3} \end{align*} of the original number of students are left in the room. We need to find how many students were in the room at the start. Step 2 Translate into a mathematical equation. Initially we have an unknown number of students in the room. We can refer to them as the original number. Let’s define the variable \begin{align*} x= \end{align*} the original number of students in the room. 25 students leave the room. The number of students left in the room is: Text Translates to Mathematical Expression the original number of students in the room \begin{align*} \rightarrow\end{align*} \begin{align*} x\end{align*} 25 students leave the room \begin{align*} \rightarrow\end{align*} \begin{align*} x-25 \end{align*} \begin{align*}\frac{2}{3}\end{align*} of the original number is left in the room \begin{align*} \rightarrow\end{align*} \begin{align*} \frac{2}{3} x \end{align*} \begin{align*}x-25=\frac{2}{3}x & & \text{THIS IS OUR EQUATION}.\end{align*} Step 3 Solve the equation. Add 25 to both sides. \begin{align*} x-25& = \frac{2}{3}x\\ x-25+25& = \frac{2}{3}x+25\\ x& = \frac{2}{3}x+25\end{align*} Subtract \begin{align*} \frac{2}{3}x \end{align*} from both sides. \begin{align*} x-\frac{2}{3}x & = \frac{2}{3}x-\frac{2}{3}x+25\\ \frac{1}{3}x& = 25\end{align*} Multiply both sides by 3. \begin{align*}3\cdot \frac{1}{3}x& = 25\cdot 3\\ x&= 75\end{align*} Remember that \begin{align*} x \end{align*} represents the original number of students in the room. So, Answer There were 75 students in the room to start with. Step 4 Check the answer: If we start with 75 students in the room and 25 of them leave, then there are \begin{align*} 75-25=50 \end{align*} students left in the room. \begin{align*} \frac{2} {3}\end{align*} of the original number is \begin{align*} \frac{2} {3} \cdot 75 = 50\end{align*} This means that the number of students who are left over equals to \begin{align*} \frac{2} {3} \end{align*} of the original number. The answer checks out. The method of defining variables and writing a mathematical equation is the method you will use the most in an algebra course. This method is often used together with other techniques such as making a table of values, creating a graph, drawing a diagram and looking for a pattern. ## Review Questions Day Profit 1 20 2 40 3 60 4 80 5 100 1. Write a mathematical equation that describes the relationship between the variables in the table: 2. what is the profit on day 10? 1. Write a mathematical equation that describes the situation: A full cookie jar has 24 cookies. How many cookies are left in the jar after you have eaten some? 2. How many cookies are in the jar after you have eaten 9 cookies? 1. Write a mathematical equation for the following situations and solve. 1. Seven times a number is 35. What is the number? 2. One number is 25 more than 2 times another number. If each number is multiplied by five, their sum would be 350. What are the numbers? 3. The sum of two consecutive integers is 35. What are the numbers? 4. Peter is three times as old as he was six years ago. How old is Peter? 2. How much water should be added to one liter of pure alcohol to make a mixture of 25% alcohol? 3. Mia drove to Javier’s house at 40 miles per hour. Javier’s house is 20 miles away. Mia arrived at Javier’s house at 2:00 pm. What time did she leave? 4. The price of an mp3 player decreased by 20% from last year to this year. This year the price of the Player is120. What was the price last year?

1. \begin{align*} P=20t; P=\end{align*} profit; \begin{align*} t=\end{align*} number of days. \begin{align*}P =\end{align*} profit; \begin{align*}t =\end{align*} number of days
2. Profit = 200
1. \begin{align*} y=24-x; y= \end{align*} number of cookies in the jar; \begin{align*} x= \end{align*} number of cookies eaten
1. \begin{align*} x= \end{align*} the number; \begin{align*}7x=35 \end{align*}; number = 5
2. \begin{align*} x= \end{align*} another number; \begin{align*} 2x+25= \end{align*} another number; \begin{align*}5x+5(2x+25)=350\end{align*}; numbers = 15 and 55
3. \begin{align*} x =\end{align*} first integer; \begin{align*} x+1= \end{align*} second integer; \begin{align*} x+x+1=35 \end{align*} ; first integer = 17, second integer = 18
4. \begin{align*} x= \end{align*} Peter’s age; \begin{align*} x=3(x-6) \end{align*} ; Peter is 9 years old.
1. 3 liters
2. 1:30 pm
3. \$150

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