Steply

Solve x - 1 < -3 or x + 2 > 6

Step by step, with the reason for every step · Compound inequalities with and / or · Algebra 1

Step by step

The question

Question: solve x − 1 < −3 or x + 2 > 6.

"Or" means one of the parts is enough.

x − 1 < −3 or x + 2 > 6
or: one is enough

1 Solve the first part

First part: x − 1 < −3.

x − 1 < −3

We want x by itself.

The − 1 is in the way.

To remove − 1, we add 1.

We do it on both sides, so the sign stays true.

x − 1 + 1 < −3 + 1

− 1 and + 1 make zero.

Cross them out.

−3 + 1 = −2.

−3 + 1 = −2

So x is less than −2.

x < −2

2 Solve the second part

Second part: x + 2 > 6.

x + 2 > 6

We want x by itself.

The + 2 is in the way.

To remove + 2, we take away 2.

We do it on both sides, so the sign stays true.

x + 2 − 2 > 6 − 2

+ 2 and − 2 make zero.

Cross them out.

6 − 2 = 4.

6 − 2 = 4

So x is greater than 4.

x > 4

3 Show each part on a number line

4 Put the parts together

"Or": we keep every number that is shaded on either line.

So x < −2 or x > 4.

= x < −2 or x > 4

Careful!

Many people write 4 < x < −2.

But a chain means between, and no number is between like that.

So we keep the word "or".

Careful: not 4 < x < −2 ✗

Books also write the answer like this.

It is called interval notation.

A round bracket means the end is not in, like the open circle.

∞ means it goes on for ever.

∪ means or.

interval: (−∞, −2) ∪ (4, ∞)

5 Check

Let's test one number from each piece.

First −3.

First: −3 − 1 = −4

−4 is less than −3.

True!

−4 < −3

Now 5, from the other piece.

Second: 5 + 2 = 7

7 is greater than 6.

True!

So the answer works.

7 > 6

Let's look back.

First: solve the first part.

Then: solve the second part.

Then: show each part on a number line.

Then: put the parts together.

Last: check.

Answer
Careful: not 4 < x < −2 ✗

Your turn

Try this one yourself, then check:

Solve x − 1 < −4 or x + 2 > 3
Show the answer
x < −3 or x > 1

More from Compound inequalities with and / or

The whole lesson: Compound inequalities with and / or →