# Open Resources for Community College Algebra

## Section2.4Special Solution Sets

Most of the time after solving a linear equation in one variable, you end with a direct statement of what the variable must equal. For example, $$2x+8=14$$ concludes with $$x=3\text{.}$$ There is only one solution, $$3$$ in this case.
Similarly, after solving a linear inequality, you typically end with a statement like $$x\lt5\text{,}$$ and the solution set is represented with either interval notation like $$(-\infty,5)\text{,}$$ or with set-builder notation like $$\{x\mid x\lt5\}\text{.}$$ Visually on a number line, it is either the left half or the right half of the number line.
Occasionally you will have a linear equation or inequality that doesnât work this way. A linear equation might have many solutions, not just one. Or it might have none at all. An inequality might also have no solutions. Or maybe every number on the entire number line will be a solution. In this section we explore these kinds of equations and inequalities, where the solution set is âspecialâ.

### Subsection2.4.1Special Solution Sets

Recall that for the equation $$x+2=5\text{,}$$ there is only one number which will make the equation true: $$3\text{.}$$ This means that the solution is $$3\text{,}$$ and we write the solution set as $$\{3\}\text{.}$$ We say the equationâs solution set has one element, $$3\text{.}$$
Weâll now explore equations where all real numbers are solutions, and other equations where no real number is a solution.

#### Example2.4.2.

Solve for $$x$$ in $$3x=3x+4\text{.}$$
To solve this equation, we need to move all terms containing $$x$$ to one side of the equals sign. We set out to remove the $$3x$$ from the right side of the equation, by subtracting $$3x$$ from each side:
\begin{align*} 3x\amp=3x+4\\ 3x\subtractright{3x}\amp=3x+4\subtractright{3x}\\ 0\amp=4 \end{align*}
But now $$x$$ is no longer present in the equation. Our effort to remove $$3x$$ from the right side coincidentally removed it from the left side too. What value can we substitute in for $$x$$ to make $$0=4$$ become true? Thatâs a trick question; you could never make $$0=4$$ become true. We say this equation has âno solutionâ. Or we say that the equation has an âempty solution setâ. We can write an empty solution set as $$\emptyset\text{,}$$ or $$\{\text{ }\}\text{.}$$ (Be careful notto confuse the empty set symbol $$\emptyset$$ with the number zero, $$0\text{.}$$)
The equation $$0=4$$ isunambiguously false no matter what $$x$$ might be. This shows us there is no solution to the original equation.

#### Example2.4.3.

Solve for $$x$$ in $$2x+1=2x+1\text{.}$$
We will move all terms containing $$x$$ to one side of the equals sign:
\begin{align*} 2x+1\amp=2x+1\\ 2x+1\subtractright{2x}\amp=2x+1\subtractright{2x}\\ 1\amp=1 \end{align*}
Once again, $$x$$ is no longer present in the equation. What value can we substitute in for $$x$$ to make $$1=1$$ be true? This is another trick question; any value for $$x$$ would work. This means that all real numbers are solutions to the equation $$2x+1=2x+1\text{.}$$ We say this equationâs solution set contains all real numbers. We can write this set using interval notation as $$(-\infty,\infty)\text{,}$$ or use $$\mathbb{R}$$ as an abbreviation for the set of all real numbers.
The equation $$0=4$$ isunambiguously true no matter what $$x$$ might be. This shows us that all real numbers are solutions to the original linear equation.

#### Remark2.4.4.

What would have happened if we had continued the solving process after we obtained $$1=1$$ in ExampleÂ 3?
\begin{align*} 1\amp=1\\ 1\subtractright{1}\amp=1\subtractright{1}\\ 0\amp=0 \end{align*}
All we found was another unambiguously true equation, and we still conclude that all real numbers are solutions.

#### Warning2.4.5.

Note that there is a very important difference when our process ends with $$0=0\text{,}$$ compared to when it ends with $$x=0\text{.}$$ The first equation is true for all values that $$x$$ might have, and the solution set is $$(-\infty,\infty)\text{.}$$ The second situation has only one solution, $$0\text{,}$$ and the solution set is $$\{0\}\text{.}$$

#### Example2.4.6.

Solve for $$t$$ in the inequality $$4t+5\gt 4t+2\text{.}$$
To solve for $$t\text{,}$$ we aim to eliminate that $$4t$$ on the right side by subtracting $$4t$$ from each side:
\begin{align*} 4t+5\amp\gt 4t+2\\ 4t+5\subtractright{4t}\amp\gt 4t+2\subtractright{4t}\\ 5\amp\gt 2 \end{align*}
We again find ourselves with the variable completely eliminated from both sides. What values of $$t$$ would make the inequality $$5\gt 2$$ true? The answer is that all values of $$t$$ make $$5\gt 2\text{,}$$ which we know is a strange sounding sentence. So our solution set is all real numbers, which we can write as $$(-\infty,\infty)$$ or $$\mathbb{R}\text{.}$$

#### Example2.4.7.

Solve for $$x$$ in the inequality $$-5x+1\le -5x\text{.}$$
To solve for $$x\text{,}$$ we aim to eliminate the $$5x$$ from the right side by adding $$5x$$ to each side:
And yet again, the variable has gone missing. We can ask ourselves, âFor which values of $$x$$ is $$1\le 0$$ true?â The answer is that this is impossible. There is no solution to this inequality. We can write the solution set using $$\emptyset$$ or just say that there are âno solutionsâ.
Letâs summarize these two special cases that sometimes arise when solving linear equations and inequalities.

### Subsection2.4.2Further Examples

These examples may have no solutions or all real numbers as the solution set. Each

#### Example2.4.9.

calls back to the fundamental steps for solving linear equations and inequalities that we learned in SectionÂ 1, SectionÂ 2, and SectionÂ 3.

#### Checkpoint2.4.10.

Solve for $$a$$ in $$\frac{2}{3}(a+1)-\frac{5}{6}=\frac{2}{3}a\text{.}$$
Explanation.
We recall the technique from SectionÂ 3 where we clear denominators by multiplying each side of the equation by the least common denominator. Here, we will multiply each side by $$6\text{.}$$ After that, weâll be able to simplify each side of the equation and continue:
\begin{equation*} \begin{aligned} \frac{2}{3}(a+1)-\frac{5}{6}\amp=\frac{2}{3}a\\ \multiplyleft{6}\left(\frac{2}{3}(a+1)-\frac{5}{6}\right)\amp=\multiplyleft{6}\frac{2}{3}a\\ \multiplyleft{6}\frac{2}{3}(a+1)-\multiplyleft{6}\frac{5}{6}\amp=\multiplyleft{6}\frac{2}{3}a\\ 4(a+1)-5\amp=4a\\ 4a+4-5\amp=4a\\ 4a-1\amp=4a\\ 4a-1\subtractright{4a}\amp=4a\subtractright{4a}\\ -1\amp=0 \end{aligned} \end{equation*}
The statement $$-1=0$$ is unambiguously false, so the equation has no solution.

#### Checkpoint2.4.11.

Solve for $$x$$ in the equation $$3(x+2)-8=(5x+4)-2(x+1)\text{.}$$
Explanation.
A lot could be simplified on each side before continuing, by distributing and combining like terms.
\begin{equation*} \begin{aligned} 3(x+2)-8\amp=(5x+4)-2(x+1)\\ 3x+6-8\amp=5x+4-2x-2\\ 3x-2\amp=3x+2 \end{aligned} \end{equation*}
From here, we subtract $$3x$$ from each side:
\begin{equation*} \begin{aligned} 3x-2\subtractright{3x}\amp=3x+2\subtractright{3x}\\ -2\amp=2 \end{aligned} \end{equation*}
As the equation $$-2=2$$ is outright false, there is no solution to this equation.

#### Checkpoint2.4.12.

Solve for $$z$$ in the inequality $$\frac{3z}{5}+\frac{1}{2}\le\left(\frac{z}{10}+\frac{3}{4}\right)+\left(\frac{z}{2}-\frac{1}{4}\right)\text{.}$$
Explanation.
We start by multiplying each side of the inequality by the LCD, which is $$20\text{.}$$ Then we can continue:
\begin{equation*} \begin{aligned} \frac{3z}{5}+\frac{1}{2}\amp\le\left(\frac{z}{10}+\frac{3}{4}\right)+\left(\frac{z}{2}-\frac{1}{4}\right)\\ \multiplyleft{20}\left(\frac{3z}{5}+\frac{1}{2}\right)\amp\le \multiplyleft{20}\left(\left(\frac{z}{10}+\frac{3}{4}\right)+\left(\frac{z}{2}-\frac{1}{4}\right)\right)\\ \multiplyleft{20}\left(\frac{3z}{5}\right)+\multiplyleft{20}\left(\frac{1}{2}\right)\amp\le \multiplyleft{20}\left(\frac{z}{10}+\frac{3}{4}\right)+\multiplyleft{20}\left(\frac{z}{2}-\frac{1}{4}\right)\\ 20\cdot\left(\frac{3z}{5}\right)+20\cdot\left(\frac{1}{2}\right)\amp\le \multiplyleft{20}\left(\frac{z}{10}\right)+\multiplyleft{20}\left(\frac{3}{4}\right)+\multiplyleft{20}\left(\frac{z}{2}\right)-\multiplyleft{20}\left(\frac{1}{4}\right)\\ 12z+10\amp\le 2z+15+10z-5\\ 12z+10\amp\le 12z+10\\ 12z+10\subtractright{12z}\amp\le 12z+10\subtractright{12z}\\ 10\amp\le10 \end{aligned} \end{equation*}
As the equation $$10\le10$$ is true for all values of $$z\text{,}$$ all real numbers are solutions to the original inequality. So the solution set is $$(-\infty,\infty)\text{,}$$ or just $$\mathbb{R}\text{.}$$

#### 1.

With a linear equation in one variable, what are the possibilities for how many solutions it could have? One solution? Two solutions? Are there other possibilities?

#### 2.

How will you know when a linear equation or inequality has no solution?

#### 3.

How will you know when all numbers are solutions to a linear equation or inequality?

### Exercises2.4.4Exercises

#### Notation

##### 1.
What is one valid way to communicate that there are no solutions to an equation? (There are several correct answers.)
##### 2.
What is one valid way to communicate that all real numbers are solutions to an equation? (There are several correct answers.)

#### Skills Practice

##### Equations with Special Solution Sets.
Solve the equation for its variable.
###### 3.
$${8t}={8t-2}$$
###### 4.
$${9P}={9P-9}$$
###### 5.
$${9k+4}={9k+4}$$
###### 6.
$${2F-3}={2F-3}$$
###### 7.
$${-6D-11+9D}={-9+3D-2}$$
###### 8.
$${-4C-9+3C}={-1-C-8}$$
###### 9.
$${-2W+1-4W}={8-6W+6}$$
###### 10.
$${v-1+9v}={-5+10v+2}$$
###### 11.
$${3\mathopen{}\left(n+2\right)}={3\mathopen{}\left(n+5\right)}$$
###### 12.
$${5\mathopen{}\left(Z-5\right)}={5\mathopen{}\left(Z-7\right)}$$
###### 13.
$${5\mathopen{}\left(X+4\right)-\left(13X-1\right)}={-8+4\mathopen{}\left(7-2X\right)}$$
###### 14.
$${5\mathopen{}\left(s-1\right)-\left(13s+1\right)}={-12-2\mathopen{}\left(-3+4s\right)}$$
###### 15.
$${5-5\mathopen{}\left(2-3t\right)}={17t-\left(5+2t\right)}$$
###### 16.
$${29-4\mathopen{}\left(8+5x\right)}={-17x-\left(4+3x\right)}$$
###### 17.
$${{\frac{3}{7}}E}={{\frac{3}{7}}E-5}$$
###### 18.
$${{\frac{4}{9}}G}={{\frac{4}{9}}G-6}$$
###### 19.
$${{\frac{7}{6}}T-10+{\frac{1}{5}}T}={-9+{\frac{41}{30}}T-1}$$
###### 20.
$${{\frac{4}{7}}u+3+{\frac{4}{5}}u}={6+{\frac{48}{35}}u-3}$$
###### 21.
$${{\frac{1}{8}}\mathopen{}\left(X-2\right)-\left({\frac{131}{24}}X+7\right)}={-{\frac{97}{4}}+{\frac{8}{3}}\mathopen{}\left(6-2X\right)}$$
###### 22.
$${{\frac{7}{9}}\mathopen{}\left(m-2\right)-\left({\frac{61}{9}}m-3\right)}={{\frac{121}{9}}+{\frac{3}{2}}\mathopen{}\left(-8-4m\right)}$$
##### Inequalities with Special Solution Sets.
Solve the inequality for its variable.
###### 23.
$${9K}\geq{9K-1}$$
###### 24.
$${2g}\leq{2g-8}$$
###### 25.
$${3C+4}\gt{4+3C}$$
###### 26.
$${4Y-2}\lt{4Y-2}$$
###### 27.
$${-2q+6-9q}\geq{-1-11q+7}$$
###### 28.
$${p+5+4p}\gt{6+5p-1}$$
###### 29.
$${3S-2-4S}\leq{-6-S-3}$$
###### 30.
$${5s-5+9s}\leq{2+14s-9}$$
###### 31.
$${7\mathopen{}\left(K+3\right)}\gt{7\mathopen{}\left(K-9\right)}$$
###### 32.
$${9\mathopen{}\left(x-4\right)}\lt{9\mathopen{}\left(x-1\right)}$$
###### 33.
$${7-5\mathopen{}\left(2+5Z\right)}\lt{-21Z-\left(3+4Z\right)}$$
###### 34.
$${35-4\mathopen{}\left(9+3d\right)}\gt{-15d-\left(2-3d\right)}$$
###### 35.
$${{\frac{3}{5}}K}\leq{{\frac{3}{5}}K-5}$$
###### 36.
$${{\frac{4}{7}}g}\gt{{\frac{4}{7}}g+8}$$
###### 37.
$${{\frac{5}{6}}R-13+{\frac{6}{5}}R}\gt{-8+{\frac{61}{30}}R-5}$$
###### 38.
$${{\frac{2}{7}}r-1+{\frac{9}{4}}r}\gt{6+{\frac{71}{28}}r-7}$$
###### 39.
$${{\frac{8}{7}}\mathopen{}\left(v-6\right)-\left({\frac{128}{63}}v+6\right)}\gt{-{\frac{649}{63}}+{\frac{4}{9}}\mathopen{}\left(1-2v\right)}$$
###### 40.
$${{\frac{5}{9}}\mathopen{}\left(K-6\right)-\left({\frac{167}{9}}K-2\right)}\geq{-{\frac{89}{6}}+{\frac{9}{2}}\mathopen{}\left(3-4K\right)}$$

#### Challenge

##### 41.
Fill in the right side of the equation to create a linear equation that meets the description.
1. Create a linear equation with solution set $$\{2\}\text{.}$$
$$20(x + 4) =$$
2. Create a linear equation with infinitely many solutions.
$$20(x + 4) =$$