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Solve 3x - 4 < 5 or 2x + 1 > 11

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

Step by step

The question

Question: solve 3x − 4 < 5 or 2x + 1 > 11.

"Or" means one of the parts is enough.

3x − 4 < 5 or 2x + 1 > 11
or: one is enough

1 Solve the first part

First part: 3x − 4 < 5.

3x − 4 < 5

We want x by itself.

Two things are stuck to x: times 3 and − 4.

The − 4 is on the outside, so it comes off first.

Then the times 3.

To remove − 4, we add 4.

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

3x − 4 + 4 < 5 + 4

− 4 and + 4 make zero.

Cross them out.

5 + 4 = 9.

5 + 4 = 9

Now we have 3x < 9.

3x < 9

3x means 3 times x.

We want just x.

To remove times 3, we divide by 3.

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

3x3 < 93

3 divided by 3 is 1.

So just x is left.

9 divided by 3 is 3.

93 = 3

So x is less than 3.

x < 3

2 Solve the second part

Second part: 2x + 1 > 11.

2x + 1 > 11

We want x by itself.

Two things are stuck to x: times 2 and + 1.

The + 1 is on the outside, so it comes off first.

Then the times 2.

To remove + 1, we take away 1.

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

2x + 1 − 1 > 11 − 1

+ 1 and − 1 make zero.

Cross them out.

11 − 1 = 10.

11 − 1 = 10

Now we have 2x > 10.

2x > 10

2x means 2 times x.

We want just x.

To remove times 2, we divide by 2.

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

2x2 > 102

2 divided by 2 is 1.

So just x is left.

10 divided by 2 is 5.

102 = 5

So x is greater than 5.

x > 5

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 < 3 or x > 5.

= x < 3 or x > 5

Careful!

Many people write 5 < x < 3.

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

So we keep the word "or".

Careful: not 5 < x < 3 ✗

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: (−∞, 3) ∪ (5, ∞)

5 Check

Let's test one number from each piece.

First 2.

First: 3(2) − 4 = 6 − 4 = 2

2 is less than 5.

True!

2 < 5

Now 6, from the other piece.

Second: 2(6) + 1 = 12 + 1 = 13

13 is greater than 11.

True!

So the answer works.

13 > 11

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 5 < x < 3 ✗

Your turn

Try this one yourself, then check:

Solve 3x − 4 < 14 or 2x + 1 > 17
Show the answer
x < 6 or x > 8

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