Worked examples
The question
Question: solve |x| = 6.
|x| = 6
1 Two cases
The inside is 6 away from 0.
On the number line, 6 and minus 6 are both 6 away from 0.
So the inside is 6, or minus 6.
|A| = 6 → A = 6 or A = −6
6 and −6 are both 6 away from 0
Two small equations, side by side.
x = 6 or x = −6
Careful!
Many people solve only x = 6.
Then they lose the second answer.
Careful: not only x = 6 ✗
2 Solve both
So x = 6 or x = −6.
x = 6 or x = −6
3 Check
Let's check: put x = 6 in.
The inside is 6, and the bars make it 6.
x = 6: |6| = 6
Now put x = −6 in.
The inside is −6, and the bars make it 6.
x = −6: |(−6)| = |−6| = 6
Both give 6.
Both answers work!
Let's look back.
First: two cases.
Then: solve both.
Last: check.
▶ Watch it solved
The question
Question: solve |2x + 1| = 7.
|2x + 1| = 7
1 Two cases
The inside is 7 away from 0.
On the number line, 7 and minus 7 are both 7 away from 0.
So the inside is 7, or minus 7.
|A| = 7 → A = 7 or A = −7
7 and −7 are both 7 away from 0
Two small equations, side by side.
2x + 1 = 7 or 2x + 1 = −7
Careful!
Many people solve only 2x + 1 = 7.
Then they lose the second answer.
Careful: not only 2x + 1 = 7 ✗
2 Solve both
In each one, we take away 1 on both sides.
2x + 1 − 1 = 7 − 1 or 2x + 1 − 1 = −7 − 1
+ 1 and − 1 make zero, in both.
7 minus 1 is 6.
−7 minus 1 is −8.
2x = 6 or 2x = −8
To remove times 2, we divide by 2.
x = 62 or x = −82
So x = 3 or x = −4.
x = 3 or x = −4
3 Check
Let's check: put x = 3 in.
The inside is 7, and the bars make it 7.
x = 3: |2(3) + 1| = |7| = 7
Now put x = −4 in.
The inside is −7, and the bars make it 7.
x = −4: |2(−4) + 1| = |−7| = 7
Both give 7.
Both answers work!
Let's look back.
First: two cases.
Then: solve both.
Last: check.
▶ Watch it solved
The question
Question: solve −2|x + 3| + 10 = 4.
−2|x + 3| + 10 = 4
1 Get the bars alone
We want the bars by themselves.
The + 10 is in the way, so we take away 10 on both sides.
−2|x + 3| + 10 − 10 = 4 − 10
+ 10 and − 10 make zero.
Cross them out.
4 minus 10 is −6.
−2|x + 3| = −6
To remove times −2, we divide both sides by −2.
−2|x + 3|−2 = −6−2
−2 divided by −2 is 1.
So just the bars is left.
−6 divided by −2 is 3.
|x + 3| = 3
2 Two cases
The inside is 3 away from 0.
On the number line, 3 and minus 3 are both 3 away from 0.
So the inside is 3, or minus 3.
|A| = 3 → A = 3 or A = −3
3 and −3 are both 3 away from 0
Two small equations, side by side.
x + 3 = 3 or x + 3 = −3
Careful!
Many people solve only x + 3 = 3.
Then they lose the second answer.
Careful: not only x + 3 = 3 ✗
3 Solve both
In each one, we take away 3 on both sides.
x + 3 − 3 = 3 − 3 or x + 3 − 3 = −3 − 3
+ 3 and − 3 make zero, in both.
So x = 0 or x = −6.
x = 0 or x = −6
4 Check
Let's check: put x = 0 in.
The inside is 3, and the bars make it 3.
x = 0: −2|0 + 3| + 10 = −2|3| + 10
= −2(3) + 10 = 4
Now put x = −6 in.
The inside is −3, and the bars make it 3.
x = −6: −2|(−6) + 3| + 10 = −2|−3| + 10
= −2(3) + 10 = 4
Both give 4.
Both answers work!
Let's look back.
First: get the bars alone.
Then: two cases.
Then: solve both.
Last: check.
▶ Watch it solved
