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Section3.5Special Solution Sets

ObjectivesPCC Course Content and Outcome Guide

Most of the time, a linear equation's final equivalent equation looks like \(x=3\text{,}\) and the solution set is written to show that there is only one solution: \(\{3\}\text{.}\) Similarly, a linear inequality's final equivalent equation looks like \(x\lt5\text{,}\) and the solution set is represented with either \((-\infty,5)\) in interval notation or \(\{x|x\lt5\}\) in set-builder notation. It's possible that both linear equations and inequalities have all real numbers as possible solutions, and it's possible that no real numbers are solutions to each. In this section, we will explore these special solution sets.

Figure3.5.1Alternative Video Lessons

Subsection3.5.1Special 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 our solution is \(3\text{,}\) and we write the solution set as \(\{3\}\text{.}\) We say the equation's solution set has one element, \(3\text{.}\)

We'll now explore equations that have all real numbers as possible solutions or no real numbers as possible solutions.

Example3.5.2

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:

\begin{align*} 3x\amp=3x+4\\ 3x\subtractright{3x}\amp=3x+4\subtractright{3x}\\ 0\amp=4 \end{align*}

Notice that \(x\) is no longer present in the equation. What value can we substitute into \(x\) to make \(0=4\) true? Nothing! We say this equation has no solution. Or, the equation has an empty solution set. We can write this as \(\emptyset\text{,}\) which is the symbol for the empty set.

The equation \(0=4\) is known a false statement, as it never holds true.

Example3.5.3

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*}

At this point, \(x\) is no longer contained in the equation. What value can we substitute into \(x\) to make \(1=1\) true? Any number! This means that all real numbers are possible solutions to the equation \(2x+1=2x+1\text{.}\) We say this equation's solution set contains all real numbers. We can write this set using set-builder notation as \(\{x\mid x\text{ is a real number}\}\) or using interval notation as \((-\infty,\infty)\text{.}\)

The equation \(1=1\) is known as an identity as it is always true.

Remark3.5.4

What would have happened if we had continued solving after we obtained \(1=1\) in Example 3.5.3?

\begin{align*} 1\amp=1\\ 1\subtractright{1}\amp=1\subtractright{1}\\ 0\amp=0 \end{align*}

As we can see, all we found was another identity — a different equation that is true for all values of \(x\text{.}\)

Warning3.5.5

Note that there is a very important difference between ending with an equivalent equation of \(0=0\) and \(x=0\text{.}\) The first holds true for all real numbers, and the solution set is \(\{x\mid x\text{ is a real number}\}\text{.}\) The second has only one solution: \(0\text{.}\) We write that solution set to show that only the number zero is the solution: \(\{0\}\text{.}\)

Example3.5.6

Solve for \(t\) in the inequality \(4t+5\gt 4t+2\text{.}\)

To solve for \(t\text{,}\) we will first subtract \(4t\) from each side to get all terms containing \(t\) on one 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*}

Notice that again, the variable \(t\) is no longer contained in the inequality. We then need to consider which values of \(t\) make the inequality true. The answer is all values, so our solution set is all real numbers, which we can write as \(\{t\mid t\text{ is a real number}\}\text{.}\)

Example3.5.7

Solve for \(x\) in the inequality \(-5x+1\le -5x\text{.}\)

To solve for \(x\text{,}\) we will first add \(5x\) to each side to get all terms containing \(x\) on one side:

\begin{align*} -5x+1\amp\le -5x\\ -5x+1\addright{5x}\amp\le -5x\addright{5x}\\ 1\amp\le 0 \end{align*}

Once more, the variable \(x\) is absent. So we can ask ourselves, “For which values of \(x\) is \(1\le 0\) true?” The answer is none, and so there is no solution to this inequality. We can write the solution set using \(\emptyset\text{.}\)

Remark3.5.8

Again consider what would have happened if we had continued solving after we obtained \(1\le 0\) in Example 3.5.7.

\begin{align*} 1\amp\le 0 \\ 1\subtractright{1}\amp\le 0\subtractright{1} \\ 0\amp\le -1 \end{align*}

As we can see, all we found was another false statement—a different equation that is not true for any real number.

Let's summarize the two special cases when solving linear equations and inequalities:

All Real Numbers

When the equivalent equation or inequality is an identity such as \(2=2\) or \(0\lt2\text{,}\) all real numbers are solutions. We write this solution set as either \(\{x\mid x\text{ is a real number}\}\) or \((-\infty,\infty)\text{.}\)

No Solution

When the equivalent equation or inequality is a false statement such as \(0=2\) or \(0\gt2\text{,}\) no real number is a solution. We write this solution set as either \(\{\ \}\) or \(\emptyset\) or write the words “no solution exists.”

List3.5.9Special Solution Sets for Equations and Inequalities

Subsection3.5.2Solving Equations and Inequalities with Special Solution Sets

Let's look at a few more complicated examples.

Example3.5.10

Solve for \(a\) in \(\frac{2}{3}(a+1)-\frac{5}{6}=\frac{2}{3}a\text{.}\)

To solve this equation for \(a\text{,}\) we'll want to recall the technique of multiplying each side of the equation by the LCD of all fractions. Here, this means that we will multiply each side by \(6\) as our first step. After that, we'll be able to simplify each side of the equation and continue solving for \(a\text{:}\)

\begin{align*} \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=\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{align*}

The statement \(-1=0\) is false, so the equation has no solution. We state the empty set as the solution set: \(\emptyset\text{.}\)

Example3.5.11

Solve for \(x\) in the equation \(3(x+2)-8=(5x+4)-2(x+1)\text{.}\)

To solve for \(x\text{,}\) we will first need to simplify the left side and right side of the equation as much as possible by distributing and combining like terms:

\begin{align*} 3(x+2)-8\amp=(5x+4)-2(x+1)\\ 3x+6-8\amp=5x+4-2x-2\\ 3x-2\amp=3x+2 \end{align*}

From here, we'll want to subtract \(3x\) from each side:

\begin{align*} 3x-2\subtractright{3x}\amp=3x+2\subtractright{3x}\\ -2\amp=2 \end{align*}

As the equation \(-2=2\) is not true for any value of \(x\text{,}\) there is no solution to this equation. We write the solution set as: \(\emptyset\text{.}\)

Example3.5.12

Solve for \(z\) in the inequality \(\frac{3z}{5}+\frac{1}{2}\le\left(\frac{z}{10}+\frac{3}{4}\right)+\left(\frac{z}{2}-\frac{1}{4}\right)\text{.}\)

To solve for \(z\text{,}\) we will first need to multiply each side of the inequality by the LCD, which is \(40\text{.}\) After that, we'll finish solving by putting all terms containing a variable on one side of the inequality:

\begin{align*} \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{40}\left(\frac{3z}{5}+\frac{1}{2}\right)\amp\le \multiplyleft{40}\left(\left(\frac{z}{10}+\frac{3}{4}\right)+\left(\frac{z}{2}-\frac{1}{4}\right)\right)\\ \multiplyleft{40}\left(\frac{3z}{5}\right)+\multiplyleft{40}\left(\frac{1}{2}\right)\amp\le \multiplyleft{40}\left(\frac{z}{10}+\frac{3}{4}\right)+\multiplyleft{40}\left(\frac{z}{2}-\frac{1}{4}\right)\\ 40\cdot\left(\frac{3z}{5}\right)+40\cdot\left(\frac{1}{2}\right)\amp\le \multiplyleft{40}\left(\frac{z}{10}\right)+\multiplyleft{40}\left(\frac{3}{4}\right)+\multiplyleft{40}\left(\frac{z}{2}\right)-\multiplyleft{40}\left(\frac{1}{4}\right)\\ 24z+20\amp\le 4z+30+20z-10\\ 24z+20\amp\le 24z+20\\ 24z+20\subtractright{24z}\amp\le 24z+20\subtractright{24z}\\ 20\amp\le20 \end{align*}

As the equation \(20\le20\) is true for all values of \(z\text{,}\) all real numbers are solutions to this inequality. Thus the solution set is \(\{z\mid z\text{ is a real number}\}\text{.}\)

Subsection3.5.3Exercises

Solving Equations with Special Solution Sets

1

Solve the equation.

\(\displaystyle{ 8y=8y+8 }\)

2

Solve the equation.

\(\displaystyle{ 4r=4r+3 }\)

3

Solve the equation.

\(\displaystyle{ 2a+7=2a+7 }\)

4

Solve the equation.

\(\displaystyle{ 8b+2=8b+2 }\)

5

Solve the equation.

\(\displaystyle{ 4A-6-5A=-7-A+1 }\)

6

Solve the equation.

\(\displaystyle{ 10B-10-11B=-14-B+4 }\)

7

Solve the equation.

\(\displaystyle{ -8-7m+4 = -m+13-6m }\)

8

Solve the equation.

\(\displaystyle{ -6-9n+2 = -n+13-8n }\)

9

Solve the equation.

  1. \(10q+3=8q+3\)

  2. \(10q+3=10q+3\)

  3. \(10q+3=10q+7\)

10

Solve the equation.

  1. \(8y+7=5y+7\)

  2. \(8y+7=8y+7\)

  3. \(8y+7=8y+11\)

11

Solve the equation.

\(\displaystyle{ 2(1-4r)-(8r-4) = 17-2(4+8r) }\)

12

Solve the equation.

\(\displaystyle{ 5(5-2a)-(8a-9) = 24-2(5+9a) }\)

13

Solve the equation.

\(\displaystyle{ {40-5\!\left(9+6b\right)} = {-31b-\left(5-b\right)} }\)

14

Solve the equation.

\(\displaystyle{ {7-3\!\left(4+5A\right)} = {-16A-\left(5-A\right)} }\)

15

Solve the equation.

\(\displaystyle{ 10(B-9)=10(B-2) }\)

16

Solve the equation.

\(\displaystyle{ 6(m-5)=6(m-4) }\)

Solving Inequalities with Special Solution Sets

17

Solve this inequality. Answer using interval notation.

\(\displaystyle{ 10x > 10x+2 }\)

18

Solve this inequality. Answer using interval notation.

\(\displaystyle{ 10x > 10x+8 }\)

19

Solve this inequality. Answer using interval notation.

\(\displaystyle{ -2x \leq -2x - 6 }\)

20

Solve this inequality. Answer using interval notation.

\(\displaystyle{ -2x \leq -2x - 9 }\)

21

Solve this inequality. Answer using interval notation.

\(\displaystyle{ -3+4x+11 \geq 4x+8 }\)

22

Solve this inequality. Answer using interval notation.

\(\displaystyle{ -7+6x+12 \geq 6x+5 }\)

23

Solve this inequality. Answer using interval notation.

\(\displaystyle{ -1+6x+2 \lt 6x+1 }\)

24

Solve this inequality. Answer using interval notation.

\(\displaystyle{ -5+8x+13 \lt 8x+8 }\)

25

Solve this inequality. Answer using interval notation.

\(\displaystyle{ -9-7z+8 > -z+10-6z }\)

26

Solve this inequality. Answer using interval notation.

\(\displaystyle{ -10-5z+4 > -z+0-4z }\)

27

Solve this inequality. Answer using interval notation.

\(\displaystyle{ 5(7-8m)-(10m-5) > 16-2(4+25m) }\)

28

Solve this inequality. Answer using interval notation.

\(\displaystyle{ 2(3-2m)-(8m-5) > 18-2(9+6m) }\)

29

Solve this inequality. Answer using interval notation.

\(\displaystyle{ 2(k-10) \leq 2(k-3) }\)

30

Solve this inequality. Answer using interval notation.

\(\displaystyle{ 4(k-8) \leq 4(k-4) }\)

31

Solve this inequality. Answer using interval notation.

\(\displaystyle{ 4x \leq 4x+3 }\)

32

Solve this inequality. Answer using interval notation.

\(\displaystyle{ 6x \leq 6x+9 }\)