Solve x > -2 and x < 3 Easy

The question

Question: solve x > −2 and x < 3.

"And" means both parts must be true.

Both parts are solved already: x is by itself in each one.

1 Show each part on a number line

2 Find the numbers in both

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

So −2 < x < 3.

x is between −2 and 3.

Careful!

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

It must be on both.

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.

3 Check

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

"And": it must work in both parts.

First part.

1 is greater than −2.

True!

Now the second part.

1 is less than 3.

True!

So the answer works.

Let's look back.

First: show each part on a number line.

Then: find the numbers in both.

Last: check.

▶ Watch it solved

Solve 2x + 1 > -3 and 3x - 2 <= 7 Medium

The question

Question: solve 2x + 1 > −3 and 3x − 2 ≤ 7.

"And" means both parts must be true.

1 Solve the first part

First part: 2x + 1 > −3.

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.

+ 1 and − 1 make zero.

Cross them out.

−3 − 1 = −4.

Now we have 2x > −4.

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.

2 divided by 2 is 1.

So just x is left.

−4 divided by 2 is −2.

So x is greater than −2.

2 Solve the second part

Second part: 3x − 2 ≤ 7.

We want x by itself.

Two things are stuck to x: times 3 and − 2.

The − 2 is on the outside, so it comes off first.

Then the times 3.

To remove − 2, we add 2.

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

− 2 and + 2 make zero.

Cross them out.

7 + 2 = 9.

Now we have 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.

3 divided by 3 is 1.

So just x is left.

9 divided by 3 is 3.

So x is less than or equal to 3.

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 −2 < x ≤ 3.

x is between −2 and 3.

Careful!

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

It must be on both.

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.

A square bracket means the end is in, like the filled circle.

5 Check

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

"And": it must work in both parts.

First part.

3 is greater than −3.

True!

Now the second part.

1 is less than or equal to 7.

True!

So the answer works.

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.

▶ Watch it solved

Solve x > 1 and x > 4 Harder

The question

Question: solve x > 1 and x > 4.

"And" means both parts must be true.

Both parts are solved already: x is by itself in each one.

1 Show each part on a number line

2 Find the numbers in both

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

So x > 4.

Careful!

The answer is not between the two numbers.

Both parts must be true, so only the numbers shaded on both lines count.

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.

3 Check

Let's test one number from the answer: 5.

"And": it must work in both parts.

First part.

5 is greater than 1.

True!

Now the second part.

5 is greater than 4.

True!

So the answer works.

Let's look back.

First: show each part on a number line.

Then: find the numbers in both.

Last: check.

▶ Watch it solved