SOLVING SIMPLE LINEAR INEQUALITIES WITH INTEGER COEFFICIENTS

The only difference between a linear equation in one variable
and a linear inequality in one variable is the verb.
Instead of an ‘$\,=\,$’ sign, there is an inequality symbol:

Only one new idea is needed to solve linear inequalities:
if you multiply or divide by a negative number,
then the direction of the inequality symbol must be changed.

This idea is explored in the current section. We begin with a definition:

DEFINITION linear inequality in one variable
A linear inequality in one variable is a sentence of the form: [beautiful math coming... please be patient] $$ ax + b \lt 0\ ,\ \ a\ne 0 $$ The inequality symbol can be any of the following:   $\,\lt\,$,   $\,\gt\,$,   $\,\le\,$,   or   $\,\ge\,$.

Here is a precise statement of the tools for solving linear inequalities.
Try translating them yourself, before reading the discussion that follows.
Think:   “What do these facts tell me that I can do?”

THEOREM tools for solving inequalities
For all real numbers $\,a\,$, $\,b\,$, and $\,c\,$: [beautiful math coming... please be patient] $$ a\lt b\ \ \ \text{ is equivalent to }\ \ \ a+c \lt b+c $$ If $\,c\gt 0\,$, then: [beautiful math coming... please be patient] $$ a\lt b \ \ \ \text{ is equivalent to }\ \ \ ac \lt bc $$ If $\,c \lt 0\,$, then: [beautiful math coming... please be patient] $$ a\lt b \ \ \ \text{ is equivalent to }\ \ \ ac \gt bc $$ The inequality symbol may be [beautiful math coming... please be patient] $\,\lt\,$,   $\,\gt\,$,   $\,\le\,$,   or   $\,\ge\,$, with appropriate changes made to the equivalence statements.
TRANSLATING THE THEOREM

The first sentence,

[beautiful math coming... please be patient] $a\lt b\ \ \ \text{ is equivalent to }\ \ \ a+c \lt b+c\,$,
says that you can add (or subtract) the same number to (or from) both sides of an inequality,
and it won't change the truth of the inequality.

Here's the idea:
if [beautiful math coming... please be patient] $\,a\,$ lies to the left of $\,b\,$ on a number line,
and both numbers are translated by the same amount $\,c\,$,
then $\,a+c\,$ still lies to the left of $\,b+c\,$.

translating two numbers by the same amount on a number line

The second sentence,

[beautiful math coming... please be patient] $a\lt b \ \ \ \text{ is equivalent to }\ \ \ ac \lt bc\,$,
only holds for $\,c \gt 0\,$.
This says that you can multiply (or divide)
both sides of an inequality by the same positive number,
and it won't change the truth of the inequality.

Here's the idea. Think about the situation when [beautiful math coming... please be patient] $\,c = 2\,$.
If $\,a\,$ lies to the left of $\,b\,$ on a number line,
and we double both number's distance from $\,0\,$,
then $\,2a\,$ still lies to the left of $\,2b\,$.

scaling two numbers by the same amount on a number line

It's the third sentence where something really interesting is happening.
The third sentence,

[beautiful math coming... please be patient] $a\lt b \ \ \ \text{ is equivalent to }\ \ \ ac \gt bc\,$,
holds for $\,c\lt 0\,$.
Notice the ‘$\,\lt\,$’ symbol in the sentence ‘$\,a\lt b\,$’,
but the ‘$\,\gt\,$’ symbol in the sentence ‘$\,ac\gt bc\,$’.

This says that if you multiply (or divide)
both sides of an inequality by the same negative number,
then the direction of the inequality symbol must be changed
in order to preserve the truth of the inequality.

Let's look at an example to understand this situation.
In the sketch below, ‘$\,a\lt b\,$’ is true, because [beautiful math coming... please be patient] $\,a\,$ lies to the left of $\,b\,$.
Multiplying both sides by $\,-1\,$ sends $\,a\,$ to its opposite ($\,-a\,$) ,
and sends $\,b\,$ to its opposite ($\,-b\,$) .
Now, the opposite of $\,a\,$ is to the right of the opposite of $\,b\,$.
That is, [beautiful math coming... please be patient] $\, -a\gt -b\,$.

multiplying by -1 sends a number to its opposite

This simple idea is the reason why you must flip the inequality symbol
when multiplying or dividing by a negative number.
SO REMEMBER!
When you multiply or divide both sides of an inequality by a negative number,
then you must change the direction of the inequality symbol!

Here are a couple sketches to further help you understand this concept:


green for initial inequality
red for positions after multiplying by $\,-1$

$1 < 2$   $\ldots$   multiply both sides by $\,-1\,$ to get   $\ldots$   $-1 > -2$

$1\,$ lies to the left of $\,2\,$   $\ldots$   but   $\ldots$   $\,-1\,$ lies to the right of $\,-2$

Put your left hand on $\,1\,$.
Put your right hand on $\,2\,$.
Let both hands follow the arrows.
Your hands cross over!

green for initial inequality
red for positions after multiplying by $\,-1$

$1 > -2$   $\ldots$   multiply both sides by $\,-1\,$ to get   $\ldots$   $-1 < 2$

$1\,$ lies to the right of $\,-2\,$   $\ldots$   but   $\ldots$   $\,-1\,$ lies to the left of $\,2$

Put your right hand on $\,1\,$.
Put your left hand on $\,-2\,$.
Let both hands follow the arrows.
Your hands cross over!
EXAMPLES:
Solve: [beautiful math coming... please be patient] $-6 - 3x \ge 4$
Solution: Write a nice, clean list of equivalent sentences:
[beautiful math coming... please be patient] $-6 - 3x \ge 4$ (original sentence)
[beautiful math coming... please be patient] $-3x \ge 10$ (add $\,6\,$ to both sides)
[beautiful math coming... please be patient] $x \le -\frac{10}{3}$ (divide both sides by $\,-3\,$; change the direction of the inequality symbol)
Solve: [beautiful math coming... please be patient] $2x \lt 5$
Answer: [beautiful math coming... please be patient] $x \lt \frac{5}{2}$
Solve: [beautiful math coming... please be patient] $-3x \gt 7$
Answer: [beautiful math coming... please be patient] $x \lt -\frac{7}{3} $
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Solving Linear Inequalities with Integer Coefficients

 
 

Solve the given inequality.
Write the result in the most conventional way.

For more advanced students, a graph is displayed.
For example, the inequality $\,-6 - 3x \ge 4\,$
is optionally accompanied by the graph of $\,y = -6 - 3x\,$ (the left side of the inequality, dashed green)
and the graph of $\,y = 4\,$ (the right side of the inequality, solid purple).
In this example, you are finding the values of $\,x\,$ where the green graph lies on or above the purple graph.
Click the “show/hide graph” button if you prefer not to see the graph.

CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5 6 7
8 9 10 11 12 13 14
AVAILABLE MASTERED IN PROGRESS

Solve:
(MAX is 14; there are 14 different problem types.)