For this lesson, we will learn the similarities of Inequalities and Equations. We will learn how you can solve an Inequality the same way as an Equation, **except** for one rule simple rule.

If a > b is true, then a + c > b + c

If a > b is true and c is a positive number (i.e. c > 0), then ac > ab is true.

If a > b is true and c is a **negative** number (i.e. c < 0), then ac < bc

You can largely think of solving Inequalities with the same method as linear equations **except** when you are using the Inequality Property of Multiplication and Division when **c is negative (c < 0)**

For example, we have the true Inequality 6 < 12, watch what happens when we:

**Add 2 to both sides,**

6 + 2 < 12 + 2,

8 < 14,

We know that 8 is less than 14, so the inequality remains true, as we have followed the Inequality Property of Addition and Subtraction

**Subtract 2 from both sides**,

6 - 2 < 12 - 2,

4 < 10,

The inequality remains true as 4 is less than 10,

Remember, even though we are dealing with a negative (we are subtracting 2 or adding a -2) it is still following the Inequality Property of Addition and Subtraction,

**Multiply both sides by 3**

(6)3 < (12)3

18 < 36

The inequality remains true

Now, **multiply both sides by -1,**

(6) (-1) < (12)(-1)

-6 < -12

We can know that -6 is **not** less than- 12, so this inequality is not true,

We started with a true inequality, 6 < 12,

We applied an incorrect property, (which was multiplying or dividing by a negative and not switching the Inequality sign)

So, remember, we follow the same rules for solving an Inequality as we do for an Equation, **except when we are multiplying or dividing by a negative**.

When we are multiplying or dividing by a negative, we must “switch” the inequality symbol.
If a > b is true and c is a **negative** number (i.e. c < 0), then ac < bc

We will also introduce what is called a solution set, this is the set of values that are the solutions to an inequality, for example, the solution set of x > 2, is all of the values that are greater than 2, represented on a number line.

When writing our solution set, we write a dot above the value that the solution says it is greater than or less than. If we had x > 2, we would write our dot above the value for 2 on our number line. When it is a ≥ or ≤ inequality the dot we fill in is **solid**, when it is a > or < inequality, the dot we fill in is hollow. From that dot we draw an arrow that runs parallel to the number line over top of all the values that fit in the solution. See examples 1, 2, & 3.

We have the inequality 2x - 1 >= 3

Remembering the Inequality Property of Addition and Subtraction and of Multiplication and Division, solve the inequality.

If it helps, for now, you can almost blank out the “≥” symbol in your head, it may cause a distraction. Think of it as an “=” and solve as if it were an equation.

First, we would add 1 to both sides (The Inequality Property of Addition and Subtraction)

2x - 1 + 1 ≥ 3 + 1

2x ≥ 4

Now divide both sides by 2 (The Inequality Property of Multiplication and Division), note we are not dividing with a negative, so we do not “switch” the inequality.

(2x) / 2 ≥ (4) / 2

x ≥ 2

So, the solution is, for 2x + 1 ≥ 3, x ≥ 2.

In words, for 2x + 1 ≥ 3 is true for every value of x that is equal to or greater than 2.

Use the answer above to write the solution set to 2x + 1 ≥ 3 on a number line. (draw the number line yourself)

We have our solution x ≥ 2, so we write a **solid** dot above the 2 and then an arrow over all the values that are in the solution set.

We have the linear inequality 4 - 2x > 12

Solve the inequality by using the properties of inequalities

So, we solve as if we can solve this as if we were solving any other equation, but keeping in mind, if we multiply or divide by a negative to both sides, we must switch the inequality,

4 - 2x > 12

Subtract 4 from both sides,

(4 -2x) - 4 > (12) - 4

-2x > 8,

Now, divide by -2, hopefully we noticed this step involves multiplying or dividing by a negative, which as it is an inequality, we are dealing with means we have to switch the > to a <,

So, divide by -2 **and** switch the negative,

(-2x) / -2 < (8) / -2

x < -4

Pick any number that fits your solution and show that it is true.

So, our solution to the inequality 4 - 2x > 12 is x < -4,

In words any x value less than -4 will be true in 4 - 2x > 12,

Pick any number less than -4, lets pick -6. Now sub that into our inequality,

4 - 2(-6) > 12,

4 -(-12) > 12

4 + 12 > 12

16 > 12,

As we would expect, -6 is a solution to the inequality, as -6 is < -4

Write a number line and graph the solution set to the inequality on that line

Our solution set is all the values that are less than -4 (x < -4),

Drawing a hollow dot over -4 (as it’s a < sign) with an arrow going over all the possible solutions we get,

We have the inequality 2(x + 2) - 2(3x + 2) ≤ 2(x + 1) - 3(x + 3)

Expand and simplify both sides of the equation.

We will use the associative, commutative, and distributive properties to expand out the brackets and simplify where possible,

First let’s multiply out all of our brackets,

2(x + 2) - 2(3x + 2) ≤ 2x + 3 - 3(x + 3)

Remember to multiply out our negatives!

2(x) + 2(2) - 2(3x) - 2(2) ≤ 2x + 3 - 3(x) - 3(3)

Multiply,

2x + 4 - 6x - 4 ≤ 2x + 3 - 3x - 9

Group together our like terms,

2x - 6x + 4 - 4 ≤ 2x -3x + 3 - 9

-4x + 0 ≤ -x -6

-4x ≤ -x - 6

Solve the inequality

Now, solve, following our properties,

Add -x to both sides,

-3x ≤ -6

Divide both sides by -3, switching our ≤ to ≥, (remembering our inequality properties)

(-3x) / -3 ≥ (-6) / -3

x ≥ 2

Plot the Solution Set on a line graph

Solve, -6x -2(3x - 1) ≥ 5(x - 3) then plot the solution on a number line.

First, expand out of the brackets

-6x -2(3x) -2(-1) ≥ 5(x) - 5(3)

-6x - 6x + 2 ≥ 5x - 15

Combine like terms

-12x + 2 ≥ 5x - 15

Subtract 5x from both sides,

-12x + 2 - 5x ≥ 5x - 15 - 5x

-17x + 2 ≥ -15

Subtract 2 from both sides,

-17x - 2 ≥ -15 - 2

-17x ≥ -17

Divide both sides by -17, Switch the inequality,

-17x / -17 ≤ -17 / -17

x ≤ 1