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Unit 2: Lesson 5

Multi-step linear inequalities

Section 2.11 : Linear Inequalities

For problems 1 – 6 solve each of the following inequalities. Give the solution in both inequality and interval notations.

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Solving Inequalities Practice Problems

Now that you've studied the many examples for solving inequalities , are you ready for some practice problems?

Let's quickly recap some of the steps for solving these practice problems.

Steps for Solving Inequalities

Steps for Graphing Your Solution to the Inequality

Now you get a chance to solve a few on your own. Try each of the practice problems on your own before checking your answer below. If you get stuck, go back and review the examples problems.

Directions: Solve each inequality and graph the solution. Click here to print out number lines for your practice.

2. 4 - 6x > -15 + 1

Solving Inequalities Practice Problem

4 - 6x > -15 + 1

Solving inequalities in one variable

This problem looked tricky only because I had to first simplify the right hand side. Again, don't forget to reverse your inequality symbol when dividing by a negative number.

It's also good practice to check your answer by substituting a number that fits the solution back into the original inequality. This will verify whether or not your inequality symbol is correct.

So, how did you do? Hopefully you did great, but if not you must check out the extra practice problems in the Algebra Class E-course .

Take a look at the questions that other students have submitted:

practice problems for solving inequalities

Compound Inequalities problem

Another compound inequality problem

practice problems for solving inequalities

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Solving Inequalities

Sometimes we need to solve Inequalities like these:

Our aim is to have x (or whatever the variable is) on its own on the left of the inequality sign:

We call that "solved".

Example: x + 2 > 12

Subtract 2 from both sides:

x + 2 − 2 > 12 − 2

How to Solve

Solving inequalities is very like solving equations ... we do most of the same things ...

... but we must also pay attention to the direction of the inequality .

Some things can change the direction !

≤ becomes ≥

≥ becomes ≤

Safe Things To Do

These things do not affect the direction of the inequality:

Example: 3x < 7+3

We can simplify 7+3 without affecting the inequality:

But these things do change the direction of the inequality ("<" becomes ">" for example):

Example: 2y+7 < 12

When we swap the left and right hand sides, we must also change the direction of the inequality :

Here are the details:

Adding or Subtracting a Value

We can often solve inequalities by adding (or subtracting) a number from both sides (just as in Introduction to Algebra ), like this:

Example: x + 3 < 7

If we subtract 3 from both sides, we get:

x + 3 − 3 < 7 − 3

And that is our solution: x < 4

In other words, x can be any value less than 4.

What did we do?

And that works well for adding and subtracting , because if we add (or subtract) the same amount from both sides, it does not affect the inequality

Example: Alex has more coins than Billy. If both Alex and Billy get three more coins each, Alex will still have more coins than Billy.

What If I Solve It, But "x" Is On The Right?

No matter, just swap sides, but reverse the sign so it still "points at" the correct value!

Example: 12 < x + 5

If we subtract 5 from both sides, we get:

12 − 5 < x + 5 − 5

That is a solution!

But it is normal to put "x" on the left hand side ...

... so let us flip sides (and the inequality sign!):

Do you see how the inequality sign still "points at" the smaller value (7) ?

And that is our solution: x > 7

Note: "x" can be on the right, but people usually like to see it on the left hand side.

Multiplying or Dividing by a Value

Another thing we do is multiply or divide both sides by a value (just as in Algebra - Multiplying ).

But we need to be a bit more careful (as you will see).

Positive Values

Everything is fine if we want to multiply or divide by a positive number :

Example: 3y < 15

If we divide both sides by 3 we get:

And that is our solution: y < 5

Negative Values

Well, just look at the number line!

For example, from 3 to 7 is an increase , but from −3 to −7 is a decrease.

See how the inequality sign reverses (from < to >) ?

Let us try an example:

Example: −2y < −8

Let us divide both sides by −2 ... and reverse the inequality !

−2y /−2 > −8 /−2

And that is the correct solution: y > 4

(Note that I reversed the inequality on the same line I divided by the negative number.)

So, just remember:

When multiplying or dividing by a negative number, reverse the inequality

Multiplying or Dividing by Variables

Here is another (tricky!) example:

Example: bx < 3b

It seems easy just to divide both sides by b , which gives us:

... but wait ... if b is negative we need to reverse the inequality like this:

But we don't know if b is positive or negative, so we can't answer this one !

To help you understand, imagine replacing b with 1 or −1 in the example of bx < 3b :

The answer could be x < 3 or x > 3 and we can't choose because we don't know b .

Do not try dividing by a variable to solve an inequality (unless you know the variable is always positive, or always negative).

A Bigger Example

Example: x−3 2 < −5.

First, let us clear out the "/2" by multiplying both sides by 2.

Because we are multiplying by a positive number, the inequalities will not change.

x−3 2 ×2 < −5 ×2

Now add 3 to both sides:

x−3 + 3 < −10 + 3

And that is our solution: x < −7

Two Inequalities At Once!

How do we solve something with two inequalities at once?

Example: −2 < 6−2x 3 < 4

First, let us clear out the "/3" by multiplying each part by 3.

Because we are multiplying by a positive number, the inequalities don't change:

−6 < 6−2x < 12

−12 < −2x < 6

Now divide each part by 2 (a positive number, so again the inequalities don't change):

−6 < −x < 3

Now multiply each part by −1. Because we are multiplying by a negative number, the inequalities change direction .

And that is the solution!

But to be neat it is better to have the smaller number on the left, larger on the right. So let us swap them over (and make sure the inequalities point correctly):

Solving Inequalities

Related Pages Solving Equations Algebraic Expressions More Algebra Lessons

In these lessons, we will look at the rules, approaches, and techniques for solving inequalities.

The following figure shows how to solve two-step inequalities. Scroll down the page for more examples and solutions.

Solve Two-Step Inequalities

The rules for solving inequalities are similar to those for solving linear equations. However, there is one exception when multiplying or dividing by a negative number.

To solve an inequality, we can:

Inequalities Of The Form “x + a > b” or “x + a < b”

Example: Solve x + 7 < 15

Solution: x + 7 < 15 x + 7 – 7 < 15 – 7 x < 8

Inequalities Of The Form “x – a < b” or “x – a > b”

Example: Solve x – 6 > 14

Solution: x – 6 > 14 x – 6+ 6 > 14 + 6 x > 20

Example: Solve the inequality x – 3 + 2 < 10

Solution: x – 3 + 2 < 10 x – 1 < 10 x – 1 + 1 < 10 + 1 x < 11

Inequalities Of The Form “a – x < b” or “a – x > b”

Example: Solve the inequality 7 – x < 9

Solution: 7 – x < 9 7 – x – 7 < 9 – 7 – x < 2 x > –2 (remember to reverse the symbol when multiplying by –1)

Example: Solve the inequality 12 > 18 – y

Solution: 12 > 18 – y 18 – y < 12 18 – y – 18 < 12 –18 – y < –6 y > 6 (remember to reverse the symbol when multiplying by –1)

Inequalities Of The Form “ < b” or “ > b”

Solving linear inequalities with like terms.

If an equation has like terms, we simplify the equation and then solve it. We do the same when solving inequalities with like terms.

Example: Evaluate 3x – 8 + 2x < 12

Solution: 3x – 8 + 2x < 12 3x + 2x < 12 + 8 5x < 20 x < 4

Example: Evaluate 6x – 8 > x + 7

Solution: 6x – 8 > x + 7 6x – x > 7 + 8 5x > 15 x > 3

Example: Evaluate 2(8 – p) ≤ 3(p + 7)

Solution: 2(8 – p) ≤ 3(p + 7) 16 – 2p ≤ 3p + 21 16 – 21 ≤ 3p + 2p –5 ≤ 5p –1 ≤ p p ≥ –1 (a < b is equivalent to b > a)

An Introduction To Solving Inequalities

Solving One-Step Linear Inequalities In One Variable

The solutions to linear inequalities can be expressed several ways: using inequalities, using a graph, or using interval notation.

The steps to solve linear inequalities are the same as linear equations, except if you multiply or divide by a negative when solving for the variable, you must reverse the inequality symbol.

Example: Solve. Express the solution as an inequality, graph and interval notation. x + 4 > 7 -2x > 8 x/-2 > -1 x - 9 ≥ -12 7x > -7 x - 9 ≤ -12

Solving Two-Step Linear Inequalities In One Variable

Example: Solve. Express the solution as an inequality, graph and interval notation. 3x + 4 ≥ 10 -2x - 1 > 9 10 ≥ -3x - 2 -8 > 5x + 12

Solving Linear Inequalities

Main rule to remember: If you multiply or divide by a negative number, the inequality flips direction.

Examples of how to solve linear inequalities are shown:

Example: Solve: 3x - 6 > 8x - 7

Students learn that when solving an inequality, such as -3x is less than 12, the goal is the same as when solving an equation: to get the variable by itself on one side.

Note that when multiplying or dividing both sides of an inequality by a negative number, the direction of the inequality sign must be switched.

For example, to solve -3x is less than 12, divide both sides by -3, to get x is greater than -4.

And when graphing an inequality on a number line, less than or greater than is shown with an open dot, and less than or equal to or greater than or equal to is shown with a closed dot.

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SAT Math : Inequalities

Study concepts, example questions & explanations for sat math, all sat math resources, example questions, example question #1 : inequalities.

|12x + 3y| < 15

What is the range of values for y, expressed in terms of x?

5 + 4x < y < 5 – 4x

–5 – 4x < y < 5 – 4x

5 – 4x < y < 5 + 4x

Recall that with absolute values and "less than" inequalities, we have to hold the following:

12x + 3y > –15

Otherwise written, this is:

–15 < 12x + 3y < 15

In this form, we can solve for y. First, we have to subtract x from all 3 parts of the inequality:

–15 – 12x < 3y < 15 – 12x

Now, we have to divide each element by 3:

(–15 – 12x)/3 < y < (15 – 12x)/3

This simplifies to:

What is a possible valid value of x?

This inequality could be rewritten as:

4x + 14 > 30 OR 4x + 14 < –30

Solve each for x:

4x + 14 > 30; 4x > 16; x > 4

4x + 14 < –30; 4x < –44; x < –11

Therefore, anything between –11 and 4 (inclusive) will not work. Hence, the answer is 7.

Example Question #3 : Inequalities

Given the inequality, |2 x – 2| > 20,

what is a possible value for x ?

For this problem, we must take into account the absolute value.

First, we solve for 2 x – 2 > 20. But we must also solve for 2 x – 2 < –20 (please notice that we negate 20 and we also flip the inequality sign).

2 x – 2 < –20

Therefore, x > 11 and x < –9.

A possible value for x would be –10 since that is less than –9.

Note: the value 11 would not be a possible value for x because the inequality sign given does not include an equal sign.

Example Question #4 : Inequalities

$x$

Example Question #5 : Inequalities

$practice problems for solving inequalities$

Example Question #131 : Algebra

$\frac{a}{5}+5> 6$

Subtract 5 from both sides of the inequality:

$\frac{a}{5}> 1$

Multiply both sides by 5:

$a> 5$

Therefore only I must be true.

Example Question #7 : Inequalities

$practice problems for solving inequalities$

Solve for both x – 3 < 2 and –( x – 3) < 2.

x – 3 < 2 and – x + 3 < 2

x < 2 + 3 and – x < 2 – 3

x < 5 and – x < –1

x < 5 and x > 1

The results are x < 5 and x > 1.

Combine the two inequalities to get 1 < x < 5

Example Question #8 : Inequalities

$practice problems for solving inequalities$

Starting with the first inequality:

$practice problems for solving inequalities$

Then, our second inequality tells us that

$practice problems for solving inequalities$

Example Question #9 : Inequalities

$practice problems for solving inequalities$

Then divide both sides by negative 3. When dividing by a negative it is important to remember to change the inequality sign. In this case the sign goes from a less than to a greater than sign.

$practice problems for solving inequalities$

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