ORCCA Solving One-Step Equations

Section 1.5 Solving One-Step Equations

Objectives: PCC Course Content and Outcome Guide

MTH 60 CCOG 2.3

permalinkWe have learned how to check whether a specific number is a solution to an equation or inequality. In this section, we will begin learning how to find the solution(s) to basic equations ourselves.

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Figure 1.5.1. Alternative Video Lesson

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Subsection 1.5.1 Imagine Filling in the Blanks

permalinkLet's start with a very simple situation—so simple, that you might have success entirely in your head without writing much down. It's not exactly the algebra we hope you will learn, but the example may serve as a good warm up.

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Example 1.5.2.

A number plus $2$ is $6\text{.}$ What is that number?

You may be so familiar with basic arithmetic that you know the answer already. The algebra approach is to translate “A number plus $2$ is $6$ ” into a math statement—in this case, an equation:

x + 2 = 6

$\begin{equation*} x+2=6 \end{equation*}$

where $x$ is the number we are trying to find. And then ask what should be substituted in for $x$ to make the equation true.

Now, how do you determine what $x$ is? One valid option is to just imagine what number you could put in place of $x$ that would result in a true equation.

Would $0$ work? No, that would mean $0+2=6\text{,}$ which is false.
Would $17$ work? No, that would mean $17+2=6\text{,}$ which is false.
Would $4$ work? Yes, because $4+2=6$ is a true equation.

So one solution to $x+2=6$ is $4\text{.}$ No other numbers are going to be solutions, because when you add $2$ to something smaller than $4\text{,}$ the result is going to be smaller than $6\text{,}$ and when you add $2$ to something larger than $4\text{,}$ the result is going to be larger than $6\text{.}$

permalinkThis approach might work for you to solve very basic equations, but in general equations are going to be too complicated to solve in your head this way. So we move on to more systematic approaches.

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Subsection 1.5.2 The Basic Principle of Algebra

permalinkLet's revisit Example 1.5.2, thinking it through differently.

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Example 1.5.3.

If a number plus $2$ is $6\text{,}$ what is the number?

If a number plus $2$ is $6\text{,}$ the number is a little smaller than $6\text{,}$ and we should be able to subtract $2$ from $6$ and get that unknown number. Doing that: $6-2=4\text{.}$ Thinking things through this way, we are using the opposite operation from addition: subtraction.

permalinkLet's try this strategy with another riddle.

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Example 1.5.4.

If a number minus $2$ is $6\text{,}$ what is the number? Now we have a number a little larger than $6$ in mind. This time, if we add $2$ to $6$ we will find the unknown number. So the unknown number is $6+2=8\text{.}$

permalinkDoes this strategy work with multiplication and division?

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Example 1.5.5.

If a number times $2$ is $6\text{,}$ what is the number? The mystery number is small, since it gets multipled by $2$ to make $6\text{.}$ If we divide $6$ by $2\text{,}$ we will find the unknown number. So the unknown number is $\frac{6}{2}=3\text{.}$

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Example 1.5.6.

If a number divided by $2$ equals $6\text{,}$ what is the number? Now we must have had a larger number to start with, since cutting it in half made $6\text{.}$ If we multiply $6$ by $2\text{,}$ we find the unknown number is $6\cdot2=12\text{.}$

These examples explore an important principle for solving an equation—applying an opposite arithmetic operation. We can revisit Example 1.5.2 and more intentionally apply this strategy. If a number plus $2$ is $6\text{,}$ what is the number? As is common in algebra, we use $x$ to represent the unknown number. The question translates into the math equation

x + 2 = 6 .

$\begin{equation*} x+2=6\text{.} \end{equation*}$

Try to envision the equals sign as the middle of a balanced scale. The left side has $2$ one-pound objects and a block with unknown weight $x$ lb. Together, the weight on the left is $x+2\text{.}$ The right side has $6$ one-pound objects. Figure 1.5.7 shows the scale.

a balance scale; its left arm has one box marked x lb and two boxes marked 1 lb; its right arm has six boxes marked 1 lb; the scale is in a state of balance — Balance scale representing $x+2=6\text{.}$

a balance scale; its left arm has a box marked x lb; its right arm has four boxes marked 1 lb; the scale is in a state of balance — Balance scale representing the solution to $x+2=6\text{,}$ after taking away $2$ from each side.

To find the weight of the unknown block, we can take away $2$ one-pound blocks from each side of the scale (to keep the scale balanced). Figure 1.5.8 shows the solution.

An equation is like a balanced scale, as the two sides of the equation are equal. In the same way that we can take away $2$ lb from each side of a balanced scale, we can subtract $2$ from each side of the equation. So instead of two pictures of balance scales, we can use algebra symbols and solve the equation $x+2=6$ in the following manner:

\begin{aligned} x + 2 & = 6 & a balanced scale \\ x + 2 - 2 & = 6 - 2 & remove the same quantity from each side \\ x & = 4 & still balanced; now it tells you the solution \end{aligned}

$\begin{align*} x+2\amp=6\amp\amp\text{a balanced scale}\\ x+2\subtractright{2}\amp=6\subtractright{2}\amp\amp\text{remove the same quantity from each side}\\ x\amp=4\amp\amp\text{still balanced; now it tells you the solution} \end{align*}$

It's important to note that each line shows what is called an equivalent equation. In other words, each equation shown is algebraically equivalent to the one above it and will have exactly the same solution(s). The final equivalent equation $x=4$ tells us that the solution to the equation is $4\text{.}$ The solution set to this equation is the set that lists every solution to the equation. For this example, the solution set is $\{4\}\text{.}$

permalinkIn Figure 1.5.9, try adding or subtracting something to each side of the equation to find its solution.

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Figure 1.5.9. Allowing Variables to Vary

We have learned we can add or subtract the same number on both sides of the equals sign, just like we can add or remove the same amount of weight on a balanced scale. Can we multiply and divide the same number on both sides of the equals sign? Let's look at Example 1.5.5 again: If a number times $2$ is $6\text{,}$ what is the number? Another balance scale can help visualize this.

a balance scale; its left arm has two boxes marked x lb; its right arm has six boxes marked 1 lb; the scale is in a state of balance — Balance scale representing the equation $2x=6\text{.}$

a balance scale; its left arm has a box marked x lb; its right arm has three boxes marked 1 lb; the scale is in a state of balance — Balance scale representing the solution to $2x=6\text{,}$ after cutting each side in half.

permalinkCurrently, the scale is balanced. If we cut the weight in half on both sides, the scale should still be balanced.

We can see from the scale that $x=3$ is correct. Removing half of the weight from each side of the scale is like dividing both sides of an equation by $2\text{:}$

\begin{aligned} 2 x & = 6 \\ \frac{2 x}{2} & = \frac{6}{2} \\ x & = 3 \end{aligned}

$\begin{align*} 2x\amp=6\\ \divideunder{2x}{2}\amp=\divideunder{6}{2}\\ x\amp=3 \end{align*}$

The equivalent equation in this example is $x=3\text{,}$ which tells us that the solution to the equation is $3$ and the solution set is $\{3\}\text{.}$

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Remark 1.5.12.

Note that when we divide each side of an equation by a number, we use the fraction line in place of the division symbol. The fact that $\frac{6}{2}=6\div2$ is a reminder that the fraction line and division symbol have the same purpose. The division symbol is rarely used in later math courses.

Similarly, we could multiply each side of an equation by $2\text{,}$ just like we can keep a scale balanced if we double the weight on each side. We will summarize these properties.

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Fact 1.5.13. Properties of Equivalent Equations.

If there is an equation $\text{Left}=\text{Right}\text{,}$ we can do the following to obtain an equivalent equation.

$\text{Left}\addright{c}=\text{Right}\addright{c}$	$\text{Left}\subtractright{c}=\text{Right}\subtractright{c}$	$\text{(Left)}\multiplyright{c}=\text{(Right)}\multiplyright{c}$	$\displaystyle\divideunder{\text{Left}}{c}=\divideunder{\text{Right}}{c}$
(add the same number to each side)	(subtract the same number from each side)	(multiply each side of the equation by the same non-zero number)	(divide each side of the equation by the same non-zero number)

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Subsection 1.5.3 Solving One-Step Equations and Stating Solution Sets

Notice that when we solved equations in Subsection 1.5.2, the final equation looked like $x=\text{number}\text{,}$ where the variable $x$ is separated from other numbers and stands alone on one side of the equals sign. The goal of solving any equation is to isolate the variable in this same manner.

permalinkPutting together both strategies (applying the opposite operation and balancing equations like a scale) that we just explored, we summarize how to solve a one-step linear equation.

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Process 1.5.14. Steps to Solving Simple (One-Step) Linear Equations.

Apply: Apply the opposite operation to both sides of the equation. If a number was added to the variable, subtract that number, and vice versa. If the variable was multiplied by a number, divide by that number, and vice versa.
Check: Check the solution. This means, verify that what you think is the solution actually solves the equation. For one reason, it's human to have made a simple arithmetic mistake, and by checking you will protect yourself from this. For another reason, there are situations where solving an equation will tell you that certain numbers are possible solutions, but they do not actually solve the original equation. Checking solutions will catch these situations.
Summarize: State the solution set, or in the case of application problems, summarize the result using a complete sentence and appropriate units.

permalinkLet's look at a few examples.

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Example 1.5.15.

Solve for $y$ in the equation $7+y=3\text{.}$

Explanation

To isolate $y\text{,}$ we need to remove $7$ from the left side. Since $7$ is being added to $y\text{,}$ we need to subtract $7$ from each side of the equation.

\begin{align*} 7+y\amp=3\\ 7+y\subtractright{7}\amp=3\subtractright{7}\\ y\amp=-4 \end{align*}

We should always check the solution when we solve equations. For this problem, we will substitute $y$ in the original equation with $-4\text{:}$

\begin{align*} 7+y\amp=3\\ 7+(\substitute{-4})\amp\stackrel{?}{=}3\\ 3\amp\stackrel{\checkmark}{=}3 \end{align*}

The solution $-4$ is checked, so the solution set is $\{-4\}\text{.}$

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Checkpoint 1.5.16.

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Checkpoint 1.5.17.

Note that in solving the equation in Checkpoint 1.5.17 we found that $-5=a\text{,}$ and did not bother to write $a=-5\text{.}$ All that really matters is that we ended with a clear statement of what $a$ must be equal to.

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Example 1.5.18.

The formula for a circle's circumference is $c=\pi d\text{,}$ where $c$ stands for circumference, $d$ stands for diameter, and $\pi$ is a constant with the value of $3.1415926\ldots\text{.}$ If a circle's circumference is 12 $\pi$ ft, find this circle's diameter.

Explanation

The circumference is given as $12\pi$ feet. Approximating $\pi$ with $3.14\text{,}$ this means the circumference is approximately 37.68 ft. It is nice to have a rough understanding of how long the circumference is, but if we use $3.14$ instead of $\pi\text{,}$ we are using a slightly smaller number than $\pi\text{,}$ and the result of any calculations we do would not be as accurate. This is why we will use the symbol $\pi$ throughout solving this equation and round only at the end in the conclusion summary (if necessary).

We will substitute $c$ in the formula with $12\pi$ and solve for $d\text{:}$

\begin{align*} c\amp=\pi d\\ \substitute{12\pi}\amp=\pi d\\ \divideunder{12\pi}{\pi}\amp=\divideunder{\pi d}{\pi}\\ 12\amp=d \end{align*}

So the circle's diameter is 12 ft.

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Example 1.5.19.

Solve for $b$ in $-b=2\text{.}$

Explanation

Note that $b$ is not yet isolated as there is a negative sign in front of it. One way to solve for $b$ is to “negate” both sides:

\begin{align*} -b\amp=2\\ \negate{-b}\amp=\negate{2}\\ b\amp=-2 \end{align*}

We removed the negative sign from $-b$ by negating both sides. A second way to remove the negative sign $-1$ from $-b$ is to divide both sides by $-1\text{.}$ If you view the original $-b$ as $-1\cdot b\text{,}$ then this approach resembles the solution from Checkpoint 1.5.17.

\begin{align*} -b\amp=2\\ -1\cdot b\amp=2\\ \divideunder{-1\cdot b}{-1}\amp=\divideunder{2}{-1}\\ b\amp=-2 \end{align*}

A third way to remove the original negative sign is to use the fact that $-1\cdot(-b)=b\text{.}$ So we could multiply on each side by $-1\text{.}$

\begin{align*} -b\amp=2\\ \multiplyleft{-1}(-b)\amp=\multiplyleft{-1}(2)\\ b\amp=-2 \end{align*}

We will check the solution by substituting $b$ in the original equation with $-2\text{:}$

\begin{align*} -b\amp=2\\ -(\substitute{-2})\amp\stackrel{\checkmark}{=}2 \end{align*}

The solution $-2$ is checked and the solution set is $\{-2\}\text{.}$

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Subsection 1.5.4 Solving One-Step Equations Involving Fractions

permalinkWhen equations have fractions, solving them will make use of the same principles. You may need to use fraction arithmetic, and there may be special considerations that will make the calculations easier. So we have separated the following examples.

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Example 1.5.20.

Solve for $g$ in $\frac{2}{3}+g=\frac{1}{2}\text{.}$

Explanation

In Section 2.3, we will learn a skill to avoid fraction operations entirely in equations like this one. For now, let's solve the equation by using subtraction to isolate $g\text{:}$

\begin{align*} \frac{2}{3}+g\amp=\frac{1}{2}\\ \frac{2}{3}+g\subtractright{\frac{2}{3}}\amp=\frac{1}{2}\subtractright{\frac{2}{3}}\\ g\amp=\frac{3}{6}-\frac{4}{6}\\ g\amp=-\frac{1}{6} \end{align*}

We will check the solution by substituting $g$ in the original equation with $-\frac{1}{6}\text{:}$

\begin{align*} \frac{2}{3}+g\amp=\frac{1}{2}\\ \frac{2}{3}+\left(\substitute{-\frac{1}{6}}\right)\amp\stackrel{?}{=}\frac{1}{2}\\ \frac{4}{6}+\left(-\frac{1}{6}\right)\amp\stackrel{?}{=}\frac{1}{2}\\ \frac{3}{6}\amp\stackrel{\checkmark}{=}\frac{1}{2} \end{align*}

The solution $-\frac{1}{6}$ is checked and the solution set is $\left\{-\frac{1}{6}\right\}\text{.}$

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Checkpoint 1.5.21.

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Example 1.5.22.

Solve for $c$ in $\frac{c}{5}=4\text{.}$

Explanation

Note that the fraction line here implies division, so our variable $c$ is being divided by $5\text{.}$ The opposite operation is to multiply by $5\text{:}$

\begin{align*} \frac{c}{5}\amp=4\\ \multiplyleft{5}\frac{c}{5}\amp=\multiplyleft{5}4\\ c\amp=20 \end{align*}

We will check the solution by substituting $c$ in the original equation with $20\text{:}$

\begin{align*} \frac{c}{5}\amp=4\\ \frac{\substitute{20}}{5}\amp\stackrel{\checkmark}{=}4 \end{align*}

The solution $20$ is checked and the solution set is $\{20\}\text{.}$

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Example 1.5.23.

Solve for $d$ in $-\frac{1}{3}d=6\text{.}$

Explanation

It's true that in this example, the variable $d$ is multiplied by $-\frac{1}{3}\text{.}$ This means that dividing each side by $-\frac{1}{3}$ would be a valid strategy for solving this equation. However, dividing by a fraction could lead to human error, so consider this alternative strategy: multiply by $-3\text{.}$

\begin{align*} -\frac{1}{3}d\amp=6\\ \multiplyleft{(-3)}\left(-\frac{1}{3}d\right)\amp=\multiplyleft{(-3)}6\\ d\amp=-18 \end{align*}

If you choose to divide each side by $-\frac{1}{3}\text{,}$ that will work out as well:

\begin{align*} -\frac{1}{3}d\amp=6\\ \divideunder{-\frac{1}{3}d}{-\frac{1}{3}}\amp=\divideunder{6}{-\frac{1}{3}}\\ d\amp=\frac{6}{1}\cdot\frac{-3}{1}\amp\amp\text{Review fraction division in }\href{section-fractions-and-fraction-arithmetic.html}{\text{A.2}}\text{.}\\ d\amp=-18 \end{align*}

This gives the same solution.

We will check the solution by substituting $d$ in the original equation with $-18\text{:}$

\begin{align*} -\frac{1}{3}d\amp=6\\ -\frac{1}{3}\cdot(\substitute{-18})\amp\stackrel{?}{=}6\\ 6\amp\stackrel{\checkmark}{=}6 \end{align*}

The solution $-18$ is checked and the solution set is $\{-18\}\text{.}$

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Example 1.5.24.

Solve for $x$ in $\frac{3x}{4}=10\text{.}$

Explanation

The variable $x$ appears to have two operations that apply to it: first multiplication by $3\text{,}$ and then division by $4\text{.}$ But note that

\begin{equation*} \frac{3x}{4}=\frac{3}{4}\cdot\frac{x}{1}=\frac{3}{4}x\text{.} \end{equation*}

If we view the left side this way, we can get away with solving the equation in one step, by multiplying on each side by the reciprocal of $\frac{3}{4}\text{.}$

\begin{align*} \frac{3x}{4}\amp=10\\ \frac{3}{4}x\amp=10\\ \multiplyleft{\frac{4}{3}}\frac{3}{4}x\amp=\multiplyleft{\frac{4}{3}}10\\ x\amp=\frac{4}{3}\cdot\frac{10}{1}\\ x\amp=\frac{40}{3} \end{align*}

We will check the solution by substituting $x$ in the original equation with $\frac{40}{3}\text{:}$

\begin{align*} \frac{3x}{4}\amp=10\\ \frac{3\left(\substitute{\frac{40}{3}}\right)}{4}\amp\stackrel{?}{=}10\\ \frac{40}{4}\amp\stackrel{?}{=}10\\ 10\amp\stackrel{\checkmark}{=}10 \end{align*}

The solution $\frac{40}{3}$ is checked and the solution set is $\left\{\frac{40}{3}\right\}\text{.}$

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Checkpoint 1.5.25.

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Reading Questions 1.5.5 Reading Questions

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1.

If you imagine the equation $2x+3=11$ as a balance scale with boxes on each side, how many boxes do you imagine on the left side? How many types of boxes do you imagine on the left side?

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2.

What is the opposite operation of multiplying by a negative number?

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3.

Every time you solve an equation, there is something you should do to guarantee success. Describe what that thing is that you should do.

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Exercises 1.5.6 Exercises

$\frac{-7}{-1}$
$\frac{4}{-1}$
$\frac{150}{-150}$
$\frac{-18}{-18}$
$\frac{12}{0}$
$\frac{0}{-9}$

Solve the equation.

Solve the equation.

$\displaystyle{ {28} = {-7t} }$
$\displaystyle{ {60} = {-9x} }$

Challenge

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91.

Write a linear equation whose solution is $x = 5\text{.}$ You may not write an equation whose left side is just “ $x$ ” or whose right side is just “ $x\text{.}$ ”

There are infinitely many correct answers to this problem. Be creative. After finding an equation that works, see if you can come up with a different one that also works.

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92.

Fill in the blanks with the numbers 18 and 67 (using each number only once) to create an equation where $x$ has the greatest possible value.

${}+x={}$
${}={}$ ${}\cdot x$