Section2.3Solving One-Step Equations
¶We 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.
Subsection2.3.1Imagine Filling in the Blanks
Let'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 want you to learn, but the example may serve as a good warm-up.
Example2.3.2
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 would be to start by translating “A number plus \(2\) is \(6\)” into a math statement — in this case, an equation:
\begin{equation*} x+2=6 \end{equation*}where \(x\) is the number we are trying to find. In other words, what should be substituted in for \(x\)…
\begin{equation*} x+2=6 \end{equation*}… 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 \(0+2\stackrel{\text{no}}{=}6\text{.}\) Would \(17\) work? No, that would mean \(17+2\stackrel{\text{no}}{=}6\text{.}\) 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 or larger than \(4\text{,}\) the result is going to be smaller or larger than \(6\text{.}\)
This 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.
Subsection2.3.2The Basic Principle of Algebra
Subsubsection2.3.2.1Opposite Operations
Let's revisit Example 2.3.2, thinking it through differently.
Example2.3.3
If a number plus \(2\) is \(6\text{,}\) what is the number?
One way to solve this riddle is to use the opposite operation. If a number plus \(2\) is \(6\text{,}\) we should be able to subtract \(2\) from \(6\) and get that unknown number. So the unknown number is \(6-2=4\text{.}\)
Let's try this strategy with another riddle.
Example2.3.4
If a number minus \(2\) is \(6\text{,}\) what is the number? This time, we should be able to add \(2\) to \(6\) to get the unknown number. So the unknown number is \(6+2=8\text{.}\)
Does this strategy work with multiplication and division?
Example2.3.5
If a number times \(2\) is \(6\text{,}\) what is the number? This time, we should be able to divide \(6\) by \(2\) to get the unknown number. So the unknown number is \(6\div2=3\text{.}\)
Example2.3.6
If a number divided by \(2\) equals \(6\text{,}\) what is the number? This time, we can multiply \(6\) by \(2\text{,}\) and the unknown number is \(6\cdot2=12\text{.}\)
These examples explore an important principle for solving an equation — applying an opposite arithmetic operation.
Subsubsection2.3.2.2Balancing Equations Like a Scale
¶We can revisit Example 2.3.2 with yet another strategy.
If a number plus \(2\) is \(6\text{,}\) what is the number?
As is common in algebra, we can use \(x\) to represent the unknown number. The question translates into the math equation
\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 2.3.7 shows the scale:
To find the weight of the unknown block, we can take away \(2\) one-pound blocks from both sides of the scale (to keep the scale balanced). The following figure 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 both sides of a balanced scale, we can subtract \(2\) on both sides of the equals sign. 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{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 quanitty 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 the one below 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 \(\{4\}\text{.}\)
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 2.3.5 again: If a number times \(2\) is \(6\text{,}\) what is the number? Another balance scale can help visualize this.
Currently, the scale is balanced. If we remove half of the weight 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 on both sides of the scale is like dividing both sides of an equation by \(2\text{:}\)
\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 \(\{3\}\text{.}\)
Remark2.3.11
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 both sides 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.
Fact2.3.12Properties of Equivalent Equations
If there is an equation \(A=B\text{,}\) we can do the following to obtain an equivalent equation:
add the same number to each side: \(A+c=B+c\)
subtract the same number from each side: \(A-c=B-c\)
multiply each side of the equation by the same non-zero number: \(A\cdot c=B\cdot c\)
divide each side of the equation by the same non-zero number: \(\frac{A}{c}=\frac{B}{c}\)
With practice, you will learn when it is helpful to use each of these properties.
Subsection2.3.3Solving One-Step Equations and Stating Solution Sets
Notice that when we solved equations in Subsection 2.3.2.2, 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.
Putting together both strategies (applying the opposite operation and balancing equations like a scale) that we just explored, we will summarize how to solve a one-step linear equation.
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 a number was multiplied by the variable, 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.
Let's look at a few examples.
Example2.3.14
Solve for \(y\) in the equation \(7+y=3\text{.}\)
Solution
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 \(\{-4\}\text{.}\)
Exercise2.3.15
Exercise2.3.16
Isolating on the Right
Note that in solving the equation in Exercise 2.3.16 we found that \(-5=a\text{,}\) which is equivalent to \(a=-5\text{.}\) We did not write \(a=-5\) as an extra step though, as \(-5=a\) identified the solution.
Example2.3.17
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 and radius.
Solution
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. And since radius is half of diameter, the radius is 6 ft.
Example2.3.18
Solve for \(b\) in \(-b=2\text{.}\)
Solution
Note that the variable \(b\) has not been isolated yet as there is a negative sign in front of it. One way to solve for \(b\) is to recognize that multiplying on both sides by \(-1\) would clear away that negative sign:
\begin{align*} -b\amp=2\\ \multiplyleft{-1}(-b)\amp=\multiplyleft{-1}(2)\\ b\amp=-2 \end{align*}We removed the negative sign from \(-b\) using the fact that \(-1\cdot(-b)=b\text{.}\) 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 Exercise 2.3.16.
\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 recognize that the opposite operation of negation is negation. So negating both sides will work out too:
\begin{align*} -b\amp=2\\ \highlight{-}(-b)\amp=\highlight{-}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{?}{=}2\\ 2\amp\stackrel{\checkmark}{=}2 \end{align*}The solution \(-2\) is checked and the solution set is \(\{-2\}\text{.}\)
Subsection2.3.4Solving One-Step Equations Involving Fractions
When 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.
Example2.3.19
Solve for \(g\) in \(\frac{2}{3}+g=\frac{1}{2}\text{.}\)
Solution
In Section 3.2, 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{?}{=}\frac{1}{2}\\ \frac{1}{2}\amp\stackrel{\checkmark}{=}\frac{1}{2} \end{align*}The solution \(-\frac{1}{6}\) is checked and the solution set is \(\left\{-\frac{1}{6}\right\}\text{.}\)
Exercise2.3.20
Example2.3.21
Solve for \(c\) in \(\frac{c}{5}=4\text{.}\)
Solution
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{?}{=}4\\ 4\amp\stackrel{\checkmark}{=}4 \end{align*}The solution \(20\) is checked and the solution set is \(\{20\}\text{.}\)
Example2.3.22
Solve for \(d\) in \(-\frac{1}{3}d=6\text{.}\)
Solution
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.
Another way to be rid of the \(-\frac{1}{3}\) is to multiply by \(-3\text{.}\) Indeed, \(-\frac{1}{3}d\) is the same as \(\frac{d}{-3}\text{,}\) and when we view the expression this way, \(d\) is being divided by \(-3\text{.}\) So multiplying by \(-3\) would be the opposite operation.
\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}\\ 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 \(\{-18\}\text{.}\)
Example2.3.23
Solve for \(x\) in \(\frac{3x}{4}=10\text{.}\)
Solution
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 \(\left\{\frac{40}{3}\right\}\text{.}\)
Exercise2.3.24
Subsection2.3.5Solving One-Step Equations
Solving One-Step Equations with Addition/Subtraction
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Solving One-Step Equations with Multiplication/Division
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
Geometry Application Problems