1.3: Patterns and Equations
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 we see. The tools of algebra help us describe 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 another.
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 in by the ticket office are related mathematically, and we can write a rule to find the amount of money taken in by the ticket office.
In words, we might say “The amount of money taken in is equal to 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 in by the ticket office.
\begin{align*}& \text{Number of visitors} \qquad 1 \qquad 2 \qquad 3 \qquad 4 \qquad 5 \qquad 6 \qquad 7\\ & \text{Money taken in} \ (\$) \quad \ \ 12 \quad \ \ 24 \quad \ 36 \quad \ 48 \quad \ \ 60 \quad \ 72 \quad \ \ 84\end{align*}
Clearly, we would need a big table 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 in 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 below. Note that this graph shows a smooth line that includes non-whole number values of \begin{align*}x\end{align*} (e.g. \begin{align*}x = 2.5\end{align*}). In real life this would not make sense, because fractions of people can’t visit a park. This is an issue of domain and range, something we will talk about later.
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 amount of money taken in 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 amount of money taken in is twelve times the number of people who enter the park.
This can be translated to:
\begin{align*}\text{the amount of money taken in} = 12 \times (\text{the number of people who enter the park})\end{align*}
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 \begin{align*}x =\end{align*} the number of people who enter the theme park.
Let \begin{align*}y =\end{align*} the total amount of money taken in at the ticket office.
We now have a fourth way to describe the relationship: with an algebraic equation.
\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 over and over again. At the end of the problem, you just need to remember which quantities \begin{align*}x\end{align*} and \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 represent real quantities. For example, if we wanted to use the algebraic equation in the example above to find the money taken in for a certain number of visitors, we would substitute that number for \begin{align*}x\end{align*} and then solve the resulting equation for \begin{align*}y\end{align*}.
Example 1
A theme park charges $12 entry to visitors. Find the money taken in if 1296 people visit the park.
Let’s break the solution to this problem down into steps. This will be a useful strategy for all the problems in this lesson.
Step 1: Extract the important information.
\begin{align*}(\text{number of dollars taken in}) &= 12 \times (\text{number of visitors})\\ (\text{number of visitors}) & = 1296\end{align*}
Step 2: Translate into a mathematical equation. To do this, we pick variables to stand for the numbers.
\begin{align*}\text{Let} \ y & = \text{(number of dollars taken in)}.\\ \text{Let} \ x & = \text{(number of visitors)}.\end{align*}
\begin{align*}(\text{number of dollars taken in}) &= 12 \times (\text{number of visitors})\\ y &= 12 \times x\end{align*}
Step 3: Substitute in any known values for the variables.
\begin{align*} y &= 12 \times x\\ x & = 1296\\ & \therefore \\ y &= 12 \times 1296 \end{align*}
Step 4: Solve the equation.
\begin{align*}y = 12 \times 1296 = 15552\end{align*}
The amount of money taken in is $15552.
Step 5: 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.
\begin{align*}(\text{number of people}) = \frac{15552}{12} = 1296\end{align*}
1296 is indeed the number of people who entered the park. 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 \begin{align*}b\end{align*} when \begin{align*}a\end{align*} is 750.
\begin{align*}& a \qquad 0 \qquad \ 10 \qquad 20 \qquad 30 \qquad 40 \ \ \qquad 50\\ & b \qquad 20 \qquad 40 \qquad 60 \qquad 80 \qquad 100 \qquad 120 \end{align*}
Step 1: Extract the important information.
We can see from the table that every time \begin{align*}a\end{align*} increases by 10, \begin{align*}b\end{align*} increases by 20. However, \begin{align*}b\end{align*} is not simply twice the value of \begin{align*}a\end{align*}. We can see that when \begin{align*}a = 0, \ b = 20\end{align*}, and this gives a clue as to what rule the pattern follows. The rule linking \begin{align*}a\end{align*} and \begin{align*}b\end{align*} is:
“To find \begin{align*}b\end{align*}, double the value of \begin{align*}a\end{align*} and add 20.”
Step 2: Translate into a mathematical equation:
Text | Translates to | Mathematical Expression |
---|---|---|
“To find \begin{align*}b\end{align*}” | \begin{align*}\rightarrow \end{align*} | \begin{align*}b =\end{align*} |
“double the value of \begin{align*}a\end{align*}” | \begin{align*}\rightarrow\end{align*} | \begin{align*}2a\end{align*} |
“add 20” | \begin{align*}\rightarrow\end{align*} | + 20 |
Our equation is \begin{align*}b = 2a + 20.\end{align*}
Step 3: Solve the equation.
The original problem asks for the value of \begin{align*}b\end{align*} when \begin{align*}a\end{align*} is 750. When \begin{align*}a\end{align*} is 750, \begin{align*}b = 2a + 20\end{align*} becomes \begin{align*}b = 2(750) + 20\end{align*}. Following the order of operations, we get:
\begin{align*} b &= 2(750) + 20\\ &= 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. In either case, checking your answer is always a good idea. In this case, we can plug our answer for \begin{align*}b\end{align*} into the equation, along with the value for \begin{align*}a\end{align*}, and see what comes out. \begin{align*}1520 = 2(750) + 20\end{align*} is TRUE because both sides of the equation are equal. 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.
\begin{align*}x-\end{align*}value | \begin{align*}y-\end{align*}value |
---|---|
-2 | 10 |
0 | 0 |
2 | -10 |
4 | -20 |
6 | -30 |
Step 1: Extract the important information.
We can see from the table that \begin{align*}y\end{align*} is five times bigger than \begin{align*}x\end{align*}. The value for \begin{align*}y\end{align*} is negative when \begin{align*}x\end{align*} is positive, and it is positive when \begin{align*}x\end{align*} is negative. Here is the rule that links \begin{align*}x\end{align*} and \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*} |
Our equation is \begin{align*}y = -5x\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 plug in -2, 0, 2, 4, and 6 for \begin{align*}x\end{align*} and solve for \begin{align*}y\end{align*}:
\begin{align*}x\end{align*} | \begin{align*}y\end{align*} |
---|---|
-2 | \begin{align*}-5(-2) = 10\end{align*} |
0 | \begin{align*}-5(0) = 0\end{align*} |
2 | \begin{align*}-5(2) = -10\end{align*} |
4 | \begin{align*}-5(4) = -20\end{align*} |
6 | \begin{align*}-5(6) = -30\end{align*} |
The \begin{align*}y-\end{align*}values in this table match the ones in the earlier table. The answer checks out.
Example 4
Zarina has a $100 gift card, and she has been spending money on the card in small regular amounts. She checks the balance on the card weekly and records it in the following table.
Week Number | Balance ($) |
---|---|
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.
The balance remaining on the card is not just a constant multiple of the week number; 100 is 100 times 1, but 78 is not 100 times 2. But there is still a pattern: the balance decreases by 22 whenever the week number increases by 1. This suggests that the balance is somehow related to the amount “-22 times the week number.”
In fact, the balance equals “-22 times the week number, plus something.” To determine what that something is, we can look at the values in one row on the table—for example, the first row, where we have a balance of $100 for week number 1.
Step 2: Translate into a mathematical equation.
First, we define our variables. Let \begin{align*}n\end{align*} stand for the week number and \begin{align*}b\end{align*} for the balance.
Then we can translate our verbal expression as follows:
Text | Translates to | Mathematical Expression |
---|---|---|
Balance equals -22 times the week number, plus something. | \begin{align*}\rightarrow\end{align*} | \begin{align*}b = -22n + ?\end{align*} |
To find out what that ? represents, we can plug in the values from that first row of the table, where \begin{align*}b = 100\end{align*} and \begin{align*}n = 1\end{align*}. This gives us \begin{align*}100 = -22(1) + ?\end{align*}.
So what number gives 100 when you add -22 to it? The answer is 122, so that is the number the ? stands for. Now our final equation is:
\begin{align*}b = -22n + 122\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 = 122 - 22 = 100\\ & n = 2 \rightarrow b = -22(2) + 122 = 122 - 44 = 78\\ & n = 3 \rightarrow b = -22(3) + 122 = 122 - 66 = 56\\ & n = 4 \rightarrow b = -22(4) + 122 = 122 - 88 = 34\end{align*}
The equation perfectly reproduces the data in the table. The answer checks out.
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 this as the original number.
Let’s define the variable \begin{align*}x =\end{align*} the original number of students in the room. After 25 students leave the room, the number of students in the room is \begin{align*}x - 25\end{align*}. We also know that the number of students left is \begin{align*}\frac{2}{3}\end{align*} of \begin{align*}x\end{align*}. So we have two expressions for the number of students left, and those two expressions are equal because they represent the same number. That means our equation is:
\begin{align*}\frac{2}{3} x = x - 25\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 &= 3 \cdot 25\\ x &= 75\end{align*}
Remember that \begin{align*}x\end{align*} represents the original number of students in the room. So, 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 \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 |
- The above table depicts the profit in dollars taken in by a store each day.
- Write a mathematical equation that describes the relationship between the variables in the table.
- What is the profit on day 10?
- If the profit on a certain day is $200, what is the profit on the next day?
- 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?
- How many cookies are in the jar after you have eaten 9 cookies?
- How many cookies are in the jar after you have eaten 9 cookies and then eaten 3 more?
- Write a mathematical equation for the following situations and solve.
- Seven times a number is 35. What is the number?
- Three times a number, plus 15, is 24. What is the number?
- Twice a number is three less than five times another number. Three times the second number is 15. What are the numbers?
- One number is 25 more than 2 times another number. If each number were multiplied by five, their sum would be 350. What are the numbers?
- The sum of two consecutive integers is 35. What are the numbers?
- Peter is three times as old as he was six years ago. How old is Peter?
- How much water should be added to one liter of pure alcohol to make a mixture of 25% alcohol?
- A mixture of 50% alcohol and 50% water has 4 liters of water added to it. It is now 25% alcohol. What was the total volume of the original mixture?
- In Crystal’s silverware drawer there are twice as many spoons as forks. If Crystal adds nine forks to the drawer, there will be twice as many forks as spoons. How many forks and how many spoons are in the drawer right now?
- 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?
- Mia left Javier’s house at 6:00 pm to drive home. This time she drove 25% faster. What time did she arrive home?
- The next day, Mia took the expressway to Javier’s house. This route was 24 miles long, but she was able to drive at 60 miles per hour. How long did the trip take?
- When Mia took the same route back, traffic on the expressway was 20% slower. How long did the return trip take?
- The price of an mp3 player decreased by 20% from last year to this year. This year the price of the player is $120. What was the price last year?
- SmartCo sells deluxe widgets for $60 each, which includes the cost of manufacture plus a 20% markup. What does it cost SmartCo to manufacture each widget?
- Jae just took a math test with 20 questions, each worth an equal number of points. The test is worth 100 points total.
- Write an equation relating the number of questions Jae got right to the total score he will get on the test.
- If a score of 70 points earns a grade of \begin{align*}C-\end{align*}, how many questions would Jae need to get right to get a \begin{align*}C-\end{align*} on the test?
- If a score of 83 points earns a grade of \begin{align*}B\end{align*}, how many questions would Jae need to get right to get a \begin{align*}B\end{align*} on the test?
- Suppose Jae got a score of 60% and then was allowed to retake the test. On the retake, he got all the questions right that he got right the first time, and also got half the questions right that he got wrong the first time. What is his new score?
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