ORCCA Special Solution Sets

Section 2.4 Special Solution Sets

Objectives: PCC Course Content and Outcome Guide

MTH 60 CCOG 2.3

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,

2 x + 8 = 14

concludes with

x = 3 .

There is only one solution,

3

in this case.

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Similarly, after solving a linear inequality, you typically end with a statement like

x < 5,

and the solution set is represented with either interval notation like

(- \infty, 5),

or with set-builder notation like

{x ∣ x < 5} .

Visually on a number line, it is either the left half or the right half of the number line.

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Occasionally you will have a linear equation or inequality that doesn’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 “special”.

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Figure 2.4.1. Alternative Video Lessons

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Subsection 2.4.1 Special Solution Sets

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Recall that for the equation

x + 2 = 5,

there is only one number which will make the equation true:

3 .

This means that the solution is

3,

and we write the solution set as

{3} .

We say the equation’s solution set has one element,

3 .

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We’ll now explore equations where all real numbers are solutions, and other equations where no real number is a solution.

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Example 2.4.2.

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Solve for

x

3 x = 3 x + 4 .

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To solve this equation, we need to move all terms containing

x

to one side of the equal sign. We set out to remove the

3 x

from the right side of the equation, by subtracting

3 x

from each side:

\begin{aligned} 3 x & = 3 x + 4 \\ 3 x - 3 x & = 3 x + 4 - 3 x \\ 0 & = 4 \end{aligned}

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But now

x

is no longer present in the equation. Our effort to remove

3 x

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? That’s a trick question; you could never make

0 = 4

become true. We say this equation has “no solution”. Or we say that the equation has an “empty solution set”. We can write an empty solution set as

\emptyset,

{} .

(Be careful notto confuse the empty set symbol

\emptyset

with the number zero,

0 .

)

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The equation

0 = 4

isunambiguously false no matter what

x

might be. This shows us there is no solution to the original equation.

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Example 2.4.3.

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Solve for

x

2 x + 1 = 2 x + 1 .

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We will move all terms containing

x

to one side of the equal sign:

\begin{aligned} 2 x + 1 & = 2 x + 1 \\ 2 x + 1 - 2 x & = 2 x + 1 - 2 x \\ 1 & = 1 \end{aligned}

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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

2 x + 1 = 2 x + 1 .

We say this equation’s solution set contains all real numbers. We can write this set using interval notation as

(- \infty, \infty),

or use

R

as an abbreviation for the set of all real numbers.

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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.

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Remark 2.4.4.

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What would have happened if we had continued the solving process after we obtained

1 = 1

in Example 3?

\begin{aligned} 1 & = 1 \\ 1 - 1 & = 1 - 1 \\ 0 & = 0 \end{aligned}

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All we found was another unambiguously true equation, and we still conclude that all real numbers are solutions.

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Warning 2.4.5.

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Note that there is a very important difference when our process ends with

0 = 0,

compared to when it ends with

x = 0 .

The first equation is true for all values that

x

might have, and the solution set is

(- \infty, \infty) .

The second situation has only one solution,

0,

and the solution set is

{0} .

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Example 2.4.6.

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Solve for

t

in the inequality

4 t + 5 > 4 t + 2 .

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To solve for

t,

we aim to eliminate that

4 t

on the right side by subtracting

4 t

from each side:

\begin{aligned} 4 t + 5 & > 4 t + 2 \\ 4 t + 5 - 4 t & > 4 t + 2 - 4 t \\ 5 & > 2 \end{aligned}

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We again find ourselves with the variable completely eliminated from both sides. What values of

t

would make the inequality

5 > 2

true? The answer is that all values of

t

make

5 > 2,

which we know is a strange sounding sentence. So our solution set is all real numbers, which we can write as

(- \infty, \infty)

R .

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Example 2.4.7.

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Solve for

x

in the inequality

- 5 x + 1 \leq - 5 x .

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To solve for

x,

we aim to eliminate the

5 x

from the right side by adding

5 x

to each side:

\begin{aligned} - 5 x + 1 & \leq - 5 x \\ - 5 x + 1 + 5 x & \leq - 5 x + 5 x \\ 1 & \leq 0 \end{aligned}

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And yet again, the variable has gone missing. We can ask ourselves, “For which values of

x

1 \leq 0

true?” 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 “no solutions”.

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Let’s summarize these two special cases that sometimes arise when solving linear equations and inequalities.

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List 2.4.8. Special Solution Sets for Equations and Inequalities

All Real Numbers: 🔗
When solving an equation or inequality boils down to an unambiguously true equation or inequality such as $2 = 2$ or $0 < 2,$ then all real numbers are solutions. We write this solution set as $(- \infty, \infty)$ or $R .$
No Solution: 🔗
When solving an equation or inequality boils down to an outright false statement such as $0 = 2$ or $0 > 2,$ then no real number is a solution. We write this solution set as either ${}$ or $\emptyset$ or write in words something like “there are no solutions”.

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Subsection 2.4.2 Further Examples

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These examples may have no solutions or all real numbers as the solution set. Each example calls back to the fundamental steps for solving linear equations and inequalities that we learned in Section 1, Section 2, and Section 3.

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Checkpoint 2.4.9.

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Solve for

a

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

Explanation.

We recall the technique from Section 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 .

After that, we’ll be able to simplify each side of the equation and continue:

\begin{aligned} \frac{2}{3} (a + 1) - \frac{5}{6} & = \frac{2}{3} a \\ 6 \cdot (\frac{2}{3} (a + 1) - \frac{5}{6}) & = 6 \cdot \frac{2}{3} a \\ 6 \cdot \frac{2}{3} (a + 1) - 6 \cdot \frac{5}{6} & = 6 \cdot \frac{2}{3} a \\ 4 (a + 1) - 5 & = 4 a \\ 4 a + 4 - 5 & = 4 a \\ 4 a - 1 & = 4 a \\ 4 a - 1 - 4 a & = 4 a - 4 a \\ - 1 & = 0 \end{aligned}

The statement

- 1 = 0

is unambiguously false, so the equation has no solution.

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Checkpoint 2.4.10.

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Solve for

x

in the equation

3 (x + 2) - 8 = (5 x + 4) - 2 (x + 1) .

Explanation.

A lot could be simplified on each side before continuing, by distributing and combining like terms.

\begin{aligned} 3 (x + 2) - 8 & = (5 x + 4) - 2 (x + 1) \\ 3 x + 6 - 8 & = 5 x + 4 - 2 x - 2 \\ 3 x - 2 & = 3 x + 2 \end{aligned}

From here, we subtract

3 x

from each side:

\begin{aligned} 3 x - 2 - 3 x & = 3 x + 2 - 3 x \\ - 2 & = 2 \end{aligned}

As the equation

- 2 = 2

is outright false, there is no solution to this equation.

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Checkpoint 2.4.11.

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Solve for

z

in the inequality

\frac{3 z}{5} + \frac{1}{2} \leq (\frac{z}{10} + \frac{3}{4}) + (\frac{z}{2} - \frac{1}{4}) .

Explanation.

We start by multiplying each side of the inequality by the LCD, which is

20 .

Then we can continue:

\begin{aligned} \frac{3 z}{5} + \frac{1}{2} & \leq (\frac{z}{10} + \frac{3}{4}) + (\frac{z}{2} - \frac{1}{4}) \\ 20 \cdot (\frac{3 z}{5} + \frac{1}{2}) & \leq 20 \cdot ((\frac{z}{10} + \frac{3}{4}) + (\frac{z}{2} - \frac{1}{4})) \\ 20 \cdot (\frac{3 z}{5}) + 20 \cdot (\frac{1}{2}) & \leq 20 \cdot (\frac{z}{10} + \frac{3}{4}) + 20 \cdot (\frac{z}{2} - \frac{1}{4}) \\ 20 \cdot (\frac{3 z}{5}) + 20 \cdot (\frac{1}{2}) & \leq 20 \cdot (\frac{z}{10}) + 20 \cdot (\frac{3}{4}) + 20 \cdot (\frac{z}{2}) - 20 \cdot (\frac{1}{4}) \\ 12 z + 10 & \leq 2 z + 15 + 10 z - 5 \\ 12 z + 10 & \leq 12 z + 10 \\ 12 z + 10 - 12 z & \leq 12 z + 10 - 12 z \\ 10 & \leq 10 \end{aligned}

As the equation

10 \leq 10

is true for all values of

z,

all real numbers are solutions to the original inequality. So the solution set is

(- \infty, \infty),

or just

R .

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Reading Questions 2.4.3 Reading Questions

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1.

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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?

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