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# 3.1: One-Step Equations

Created by: CK-12

## Learning Objectives

• Solve an equation using addition.
• Solve an equation using subtraction.
• Solve an equation using multiplication.
• Solve an equation using division.

## Introduction

Nadia is buying a new mp3 player. Peter watches her pay for the player with a $100 bill. She receives$22.00 in change, and from only this information, Peter works out how much the player cost. How much was the player?

In algebra, we can solve problems like this using an equation. An equation is an algebraic expression that involves an equals sign. If we use the letter $x$ to represent the cost of the mp3 player, we can write the equation $x + 22 = 100$. This tells us that the value of the player plus the value of the change received is equal to the $100 that Nadia paid. Another way we could write the equation would be $x = 100 - 22$. This tells us that the value of the player is equal to the total amount of money Nadia paid $(100 - 22)$. This equation is mathematically equivalent to the first one, but it is easier to solve. In this chapter, we will learn how to solve for the variable in a one-variable linear equation. Linear equations are equations in which each term is either a constant, or a constant times a single variable (raised to the first power). The term linear comes from the word line, because the graph of a linear equation is always a line. We’ll start with simple problems like the one in the last example. ## Solving Equations Using Addition and Subtraction When we work with an algebraic equation, it’s important to remember that the two sides have to stay equal for the equation to stay true. We can change the equation around however we want, but whatever we do to one side of the equation, we have to do to the other side. In the introduction above, for example, we could get from the first equation to the second equation by subtracting 22 from both sides: $x + 22 &= 100\\x + 22 - 22 &= 100 - 22\\x &= 100 - 22$ Similarly, we can add numbers to each side of an equation to help solve for our unknown. Example 1 Solve $x - 3 = 9$. Solution To solve an equation for $x$, we need to isolate $x-$that is, we need to get it by itself on one side of the equals sign. Right now our $x$ has a 3 subtracted from it. To reverse this, we’ll add 3—but we must add 3 to both sides. $x - 3 &= 9\\x - 3 + 3 &= 9 + 3\\x + 0 &= 9 + 3\\x &= 12$ Example 2 Solve $z - 9.7 = -1.026$ Solution It doesn’t matter what the variable is—the solving process is the same. $z - 9.7 &= -1.026\\ z - 9.7 + 9.7 &= -1.026 + 9.7\\z &= 8.674$ Make sure you understand the addition of decimals in this example! Example 3 Solve $x + \frac{4}{7} = \frac{9}{5}$. Solution To isolate $x$, we need to subtract $\frac{4}{7}$ from both sides. $x + \frac{4}{7} &= \frac{9}{5}\\x + \frac{4}{7} - \frac{4}{7} &= \frac{9}{5} - \frac{4}{7}\\x &= \frac{9}{5} - \frac{4}{7}$ Now we have to subtract fractions, which means we need to find the LCD. Since 5 and 7 are both prime, their lowest common multiple is just their product, 35. $x &= \frac{9}{5} - \frac{4}{7}\\x &= \frac{7 \cdot 9}{7 \cdot 5} - \frac{4 \cdot 5}{7 \cdot 5}\\x &= \frac{63}{35} - \frac{20}{35}\\x &= \frac{63 - 20}{35}\\x &= \frac{43}{35}$ Make sure you’re comfortable with decimals and fractions! To master algebra, you’ll need to work with them frequently. ## Solving Equations Using Multiplication and Division Suppose you are selling pizza for$1.50 a slice and you can get eight slices out of a single pizza. How much money do you get for a single pizza? It shouldn’t take you long to figure out that you get $8 \times \1.50 = \12.00$. You solved this problem by multiplying. Here’s how to do the same thing algebraically, using $x$ to stand for the cost in dollars of the whole pizza.

Example 4

Solve $\frac{1}{8} \cdot x = 1.5$.

Our $x$ is being multiplied by one-eighth. To cancel that out and get $x$ by itself, we have to multiply by the reciprocal, 8. Don’t forget to multiply both sides of the equation.

$8 \left ( \frac{1}{8} \cdot x \right ) &= 8(1.5)\\x &= 12$

Example 5

Solve $\frac{9x}{5} = 5$.

$\frac{9x}{5}$ is equivalent to $\frac{9}{5} \cdot x$, so to cancel out that $\frac{9}{5}$, we multiply by the reciprocal, $\frac{5}{9}$.

$\frac{5}{9} \left ( \frac{9x}{5} \right ) &= \frac{5}{9}(5)\\x &= \frac{25}{9}$

Example 6

Solve $0.25x = 5.25$.

0.25 is the decimal equivalent of one fourth, so to cancel out the 0.25 factor we would multiply by 4.

$4(0.25x) &= 4(5.25)\\x &= 21$

Solving by division is another way to isolate $x$. Suppose you buy five identical candy bars, and you are charged $3.25. How much did each candy bar cost? You might just divide$3.25 by 5, but let’s see how this problem looks in algebra.

Example 7

Solve $5x = 3.25$.

To cancel the 5, we divide both sides by 5.

$\frac{5x}{5} &= \frac{3.25}{5}\\x &= 0.65$

Example 8

Solve $7x = \frac{5}{11}$.

Divide both sides by 7.

$x &= \frac{5}{11.7}\\x &= \frac{5}{77}$

Example 9

Solve $1.375x = 1.2$.

Divide by 1.375

$x &= \frac{1.2}{1.375}\\x &= 0.8 \overline{72}$

Notice the bar above the final two decimals; it means that those digits recur, or repeat. The full answer is 0.872727272727272....

To see more examples of one - and two-step equation solving, watch the Khan Academy video series starting at http://www.youtube.com/watch?v=bAerID24QJ0.

## Solve Real-World Problems Using Equations

Example 10

In the year 2017, Anne will be 45years old. In what year was Anne born?

The unknown here is the year Anne was born, so that’s our variable $x$. Here’s our equation:

$x + 45 &= 2017\\x + 45 - 45 &= 2017 - 45\\x &= 1972$

Anne was born in 1972.

Example 11

A mail order electronics company stocks a new mini DVD player and is using a balance to determine the shipping weight. Using only one-pound weights, the shipping department found that the following arrangement balances:

How much does each DVD player weigh?

Solution

Since the system balances, the total weight on each side must be equal. To write our equation, we’ll use $x$ for the weight of one DVD player, which is unknown. There are two DVD players, weighing a total of $2x$ pounds, on the left side of the balance, and on the right side are 5 1-pound weights, weighing a total of 5 pounds. So our equation is $2x = 5$. Dividing both sides by 2 gives us $x = 2.5$.

Each DVD player weighs 2.5 pounds.

Example 12

In 2004, Takeru Kobayashi of Nagano, Japan, ate 53.5 hot dogs in 12 minutes. This was 3 more hot dogs than his own previous world record, set in 2002. Calculate:

a) How many minutes it took him to eat one hot dog.

b) How many hot dogs he ate per minute.

c) What his old record was.

Solution

a) We know that the total time for 53.5 hot dogs is 12 minutes. We want to know the time for one hot dog, so that’s $x$. Our equation is $53.5x = 12$. Then we divide both sides by 53.5 to get $x = \frac{12}{53.5}$, or $x = 0.224 \ minutes$.

We can also multiply by 60 to get the time in seconds; 0.224 minutes is about 13.5 seconds. So that’s how long it took Takeru to eat one hot dog.

b) Now we’re looking for hot dogs per minute instead of minutes per hot dog. We’ll use the variable $y$ instead of $x$ this time so we don’t get the two confused. 12 minutes, times the number of hot dogs per minute, equals the total number of hot dogs, so $12y = 53.5$. Dividing both sides by 12 gives us $y = \frac{53.5}{12}$, or $y = 4.458$ hot dogs per minute.

c) We know that his new record is 53.5, and we know that’s three more than his old record. If we call his old record $z$, we can write the following equation: $z + 3 = 53.5$. Subtracting 3 from both sides gives us $z = 50.5$. So Takeru’s old record was 50.5 hot dogs in 12 minutes.

## Lesson Summary

• An equation in which each term is either a constant or the product of a constant and a single variable is a linear equation.
• We can add, subtract, multiply, or divide both sides of an equation by the same value and still have an equivalent equation.
• To solve an equation, isolate the unknown variable on one side of the equation by applying one or more arithmetic operations to both sides.

## Review Questions

1. Solve the following equations for $x$.
1. $x = 11 = 7$
2. $x - 1.1 = 3.2$
3. $7x = 21$
4. $4x = 1$
5. $\frac{5x}{12} = \frac{2}{3}$
6. $x + \frac{5}{2} = \frac{2}{3}$
7. $x - \frac{5}{6} = \frac{3}{8}$
8. $0.01x = 11$
2. Solve the following equations for the unknown variable.
1. $q - 13 = -13$
2. $z + 1.1 = 3.0001$
3. $21s = 3$
4. $t + \frac{1}{2} = \frac{1}{3}$
5. $\frac{7f}{11} = \frac{7}{11}$
6. $\frac{3}{4} = -\frac{1}{2} - y$
7. $6r = \frac{3}{8}$
8. $\frac{9b}{16} = \frac{3}{8}$
3. Peter is collecting tokens on breakfast cereal packets in order to get a model boat. In eight weeks he has collected 10 tokens. He needs 25 tokens for the boat. Write an equation and determine the following information.
1. How many more tokens he needs to collect, $n$.
2. How many tokens he collects per week, $w$.
3. How many more weeks remain until he can send off for his boat, $r$.
4. Juan has baked a cake and wants to sell it in his bakery. He is going to cut it into 12 slices and sell them individually. He wants to sell it for three times the cost of making it. The ingredients cost him $8.50, and he allowed$1.25 to cover the cost of electricity to bake it. Write equations that describe the following statements
1. The amount of money that he sells the cake for $(u)$.
2. The amount of money he charges for each slice $(c)$.
3. The total profit he makes on the cake $(w)$.
5. Jane is baking cookies for a large party. She has a recipe that will make one batch of two dozen cookies, and she decides to make five batches. To make five batches, she finds that she will need 12.5 cups of flour and 15 eggs.
1. How many cookies will she make in all?
2. How many cups of flour go into one batch?
3. How many eggs go into one batch?
4. If Jane only has a dozen eggs on hand, how many more does she need to make five batches?
5. If she doesn’t go out to get more eggs, how many batches can she make? How many cookies will that be?

## Date Created:

Feb 22, 2012

Aug 22, 2014
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