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Solve 2x < 6 and x + 1 > 0

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

Step by step

The question

Question: solve 2x < 6 and x + 1 > 0.

"And" means both parts must be true.

2x < 6 and x + 1 > 0
and: both true

1 Solve the first part

First part: 2x < 6.

2x < 6

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 < 62

2 divided by 2 is 1.

So just x is left.

6 divided by 2 is 3.

62 = 3

So x is less than 3.

x < 3

2 Solve the second part

Second part: x + 1 > 0.

x + 1 > 0

We want x by itself.

The + 1 is in the way.

To remove + 1, we take away 1.

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

x + 1 − 1 > 0 − 1

+ 1 and − 1 make zero.

Cross them out.

0 − 1 = −1.

0 − 1 = −1

So x is greater than −1.

x > −1

3 Show each part on a number line

4 Find the numbers in both

"And": we keep only the numbers that are shaded on both lines.

So −1 < x < 3.

x is between −1 and 3.

= −1 < x < 3

Careful!

A number that is shaded on only one line is not an answer.

It must be on both.

Careful: and: only where BOTH lines are shaded

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.

interval: (−1, 3)

5 Check

Let's test one number from the answer: 1.

"And": it must work in both parts.

First part.

First: 2(1) = 2

2 is less than 6.

True!

2 < 6

Now the second part.

Second: 1 + 1 = 2

2 is greater than 0.

True!

So the answer works.

2 > 0

Let's look back.

First: solve the first part.

Then: solve the second part.

Then: show each part on a number line.

Then: find the numbers in both.

Last: check.

Answer
Careful: and: only where BOTH lines are shaded

Your turn

Try this one yourself, then check:

Solve 2x < 8 and x + 1 > −3
Show the answer
−4 < x < 4

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