Worked examples
The question
Question: solve |x − 1| = |x + 3|.
|x − 1| = |x + 3|
Both bars give how far from 0.
Two numbers the same distance from 0 are equal, or opposite.
Like 3 and minus 3.
|A| = |B| → A = B or A = −B
|3| = |−3|: 3 and −3 are both 3 away from 0
Careful!
The minus is for the whole of x + 3.
Every sign in the bracket changes.
Careful: not x − 1 = −x + 3 ✗
1 Two cases
So x − 1 = x + 3, or x − 1 = −(x + 3).
We keep the bracket: the minus is for all of it.
x − 1 = x + 3 or x − 1 = −(x + 3)
2 Solve the first case
First case: x − 1 = x + 3.
x − 1 = x + 3
3 Get x on one side
There is x on both sides.
We want x on one side only.
So we remove the x from the right side.
To remove x, we take away x.
We do it on both sides, so they stay equal.
x − 1 − x = x + 3 − x
x and − x make zero.
Cross them out.
x − x = 0.
x − x = 0
−1 = 3 is false for every x.
So this case gives no x.
−1 = 3 false
4 Solve the second case
Second case: x − 1 = −(x + 3).
x − 1 = −(x + 3)
5 Open the brackets
A minus in front of a bracket changes the sign of every part inside.
x becomes −x.
3 becomes −3.
x − 1 = −x − 3
6 Get x on one side
There is x on both sides.
We want x on one side only.
So we remove the −x from the right side.
To remove −x, we add x.
We do it on both sides, so they stay equal.
x − 1 + x = −x − 3 + x
− x and + x make zero.
Cross them out.
x + x = 2x.
x + x = 2x
Now we have 2x − 1 = −3.
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.
7 Remove the − 1
To remove − 1, we add 1.
We do it on both sides, so they stay equal.
2x − 1 + 1 = −3 + 1
− 1 and + 1 make zero.
Cross them out.
−3 + 1 = −2.
−3 + 1 = −2
Now we have 2x = −2.
2x = −2
8 Remove the 2
2x means 2 times x.
We want just x.
To remove times 2, we divide by 2.
We do it on both sides, so they stay equal.
2x2 = −22
2 divided by 2 is 1.
So just x is left.
−2 divided by 2 is −1.
−22 = −1
So x = −1!
x = −1
9 Write the answers
Only one case works.
So x = −1.
x = −1
10 Check
Let's check: put x = −1 in.
The left bars give 2.
x = −1: |−1 − 1| = |−2| = 2
The right bars give 2 too.
The same!
|−1 + 3| = |2| = 2
Let's look back.
First: two cases.
Then: solve the first case.
Then: get x on one side.
Then: solve the second case.
Then: open the brackets.
Then: get x on one side.
Then: remove the − 1.
Then: remove the 2.
Then: write the answers.
Last: check.
▶ Watch it solved
The question
Question: solve |2x − 1| = |x + 4|.
|2x − 1| = |x + 4|
Both bars give how far from 0.
Two numbers the same distance from 0 are equal, or opposite.
Like 3 and minus 3.
|A| = |B| → A = B or A = −B
|3| = |−3|: 3 and −3 are both 3 away from 0
Careful!
The minus is for the whole of x + 4.
Every sign in the bracket changes.
Careful: not 2x − 1 = −x + 4 ✗
1 Two cases
So 2x − 1 = x + 4, or 2x − 1 = −(x + 4).
We keep the bracket: the minus is for all of it.
2x − 1 = x + 4 or 2x − 1 = −(x + 4)
2 Solve the first case
First case: 2x − 1 = x + 4.
2x − 1 = x + 4
3 Get x on one side
There is x on both sides.
We want x on one side only.
So we remove the x from the right side.
To remove x, we take away x.
We do it on both sides, so they stay equal.
2x − 1 − x = x + 4 − x
x and − x make zero.
Cross them out.
2x − x = x.
2x − x = x
Now we have x − 1 = 4.
x − 1 = 4
4 Remove the − 1
We want x by itself.
The − 1 is in the way.
To remove − 1, we add 1.
We do it on both sides, so they stay equal.
x − 1 + 1 = 4 + 1
− 1 and + 1 make zero.
Cross them out.
4 + 1 = 5.
4 + 1 = 5
So x = 5!
x = 5
5 Solve the second case
Second case: 2x − 1 = −(x + 4).
2x − 1 = −(x + 4)
6 Open the brackets
A minus in front of a bracket changes the sign of every part inside.
x becomes −x.
4 becomes −4.
2x − 1 = −x − 4
7 Get x on one side
There is x on both sides.
We want x on one side only.
So we remove the −x from the right side.
To remove −x, we add x.
We do it on both sides, so they stay equal.
2x − 1 + x = −x − 4 + x
− x and + x make zero.
Cross them out.
2x + x = 3x.
2x + x = 3x
Now we have 3x − 1 = −4.
3x − 1 = −4
We want x by itself.
Two things are stuck to x: times 3 and − 1.
The − 1 is on the outside, so it comes off first.
Then the times 3.
8 Remove the − 1
To remove − 1, we add 1.
We do it on both sides, so they stay equal.
3x − 1 + 1 = −4 + 1
− 1 and + 1 make zero.
Cross them out.
−4 + 1 = −3.
−4 + 1 = −3
Now we have 3x = −3.
3x = −3
9 Remove the 3
3x means 3 times x.
We want just x.
To remove times 3, we divide by 3.
We do it on both sides, so they stay equal.
3x3 = −33
3 divided by 3 is 1.
So just x is left.
−3 divided by 3 is −1.
−33 = −1
So x = −1!
x = −1
10 Write the answers
So x = 5 or x = −1.
x = 5 or x = −1
11 Check
Let's check: put x = 5 in.
The left bars give 9.
x = 5: |2(5) − 1| = |10 − 1| = |9| = 9
The right bars give 9 too.
The same!
|5 + 4| = |9| = 9
Now put x = −1 in.
The left bars give 3.
x = −1: |2(−1) − 1| = |−2 − 1| = |−3| = 3
The right bars give 3 too.
The same!
|−1 + 4| = |3| = 3
Both answers work!
Let's look back.
First: two cases.
Then: solve the first case.
Then: get x on one side.
Then: remove the − 1.
Then: solve the second case.
Then: open the brackets.
Then: get x on one side.
Then: remove the − 1.
Then: remove the 3.
Then: write the answers.
Last: check.
▶ Watch it solved
The question
Question: solve |3x + 1| = |x − 5|.
|3x + 1| = |x − 5|
Both bars give how far from 0.
Two numbers the same distance from 0 are equal, or opposite.
Like 3 and minus 3.
|A| = |B| → A = B or A = −B
|3| = |−3|: 3 and −3 are both 3 away from 0
Careful!
The minus is for the whole of x − 5.
Every sign in the bracket changes.
Careful: not 3x + 1 = −x − 5 ✗
1 Two cases
So 3x + 1 = x − 5, or 3x + 1 = −(x − 5).
We keep the bracket: the minus is for all of it.
3x + 1 = x − 5 or 3x + 1 = −(x − 5)
2 Solve the first case
First case: 3x + 1 = x − 5.
3x + 1 = x − 5
3 Get x on one side
There is x on both sides.
We want x on one side only.
So we remove the x from the right side.
To remove x, we take away x.
We do it on both sides, so they stay equal.
3x + 1 − x = x − 5 − x
x and − x make zero.
Cross them out.
3x − x = 2x.
3x − x = 2x
Now we have 2x + 1 = −5.
2x + 1 = −5
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.
4 Remove the + 1
To remove + 1, we take away 1.
We do it on both sides, so they stay equal.
2x + 1 − 1 = −5 − 1
+ 1 and − 1 make zero.
Cross them out.
−5 − 1 = −6.
−5 − 1 = −6
Now we have 2x = −6.
2x = −6
5 Remove the 2
2x means 2 times x.
We want just x.
To remove times 2, we divide by 2.
We do it on both sides, so they stay equal.
2x2 = −62
2 divided by 2 is 1.
So just x is left.
−6 divided by 2 is −3.
−62 = −3
So x = −3!
x = −3
6 Solve the second case
Second case: 3x + 1 = −(x − 5).
3x + 1 = −(x − 5)
7 Open the brackets
A minus in front of a bracket changes the sign of every part inside.
x becomes −x.
−5 becomes +5.
3x + 1 = −x + 5
8 Get x on one side
There is x on both sides.
We want x on one side only.
So we remove the −x from the right side.
To remove −x, we add x.
We do it on both sides, so they stay equal.
3x + 1 + x = −x + 5 + x
− x and + x make zero.
Cross them out.
3x + x = 4x.
3x + x = 4x
Now we have 4x + 1 = 5.
4x + 1 = 5
We want x by itself.
Two things are stuck to x: times 4 and + 1.
The + 1 is on the outside, so it comes off first.
Then the times 4.
9 Remove the + 1
To remove + 1, we take away 1.
We do it on both sides, so they stay equal.
4x + 1 − 1 = 5 − 1
+ 1 and − 1 make zero.
Cross them out.
5 − 1 = 4.
5 − 1 = 4
Now we have 4x = 4.
4x = 4
10 Remove the 4
4x means 4 times x.
We want just x.
To remove times 4, we divide by 4.
We do it on both sides, so they stay equal.
4x4 = 44
4 divided by 4 is 1.
So just x is left.
4 divided by 4 is 1.
44 = 1
So x = 1!
x = 1
11 Write the answers
So x = −3 or x = 1.
x = −3 or x = 1
12 Check
Let's check: put x = −3 in.
The left bars give 8.
x = −3: |3(−3) + 1| = |−9 + 1| = |−8| = 8
The right bars give 8 too.
The same!
|−3 − 5| = |−8| = 8
Now put x = 1 in.
The left bars give 4.
x = 1: |3(1) + 1| = |3 + 1| = |4| = 4
The right bars give 4 too.
The same!
|1 − 5| = |−4| = 4
Both answers work!
Let's look back.
First: two cases.
Then: solve the first case.
Then: get x on one side.
Then: remove the + 1.
Then: remove the 2.
Then: solve the second case.
Then: open the brackets.
Then: get x on one side.
Then: remove the + 1.
Then: remove the 4.
Then: write the answers.
Last: check.
▶ Watch it solved
