Step by step
The question
Question: solve |a + 1| = |a − 7|.
Both bars give how far from 0.
Two numbers the same distance from 0 are equal, or opposite.
Like 3 and minus 3.
Careful!
The minus is for the whole of a − 7.
Every sign in the bracket changes.
1 Two cases
So a + 1 = a − 7, or a + 1 = −(a − 7).
We keep the bracket: the minus is for all of it.
2 Solve the first case
First case: a + 1 = a − 7.
3 Get a on one side
There is a on both sides.
We want a on one side only.
So we remove the a from the right side.
To remove a, we take away a.
We do it on both sides, so they stay equal.
a and − a make zero.
Cross them out.
a − a = 0.
1 = −7 is false for every a.
So this case gives no a.
4 Solve the second case
Second case: a + 1 = −(a − 7).
5 Open the brackets
A minus in front of a bracket changes the sign of every part inside.
a becomes −a.
−7 becomes +7.
6 Get a on one side
There is a on both sides.
We want a on one side only.
So we remove the −a from the right side.
To remove −a, we add a.
We do it on both sides, so they stay equal.
− a and + a make zero.
Cross them out.
a + a = 2a.
Now we have 2a + 1 = 7.
We want a by itself.
Two things are stuck to a: 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 take away 1.
We do it on both sides, so they stay equal.
+ 1 and − 1 make zero.
Cross them out.
7 − 1 = 6.
Now we have 2a = 6.
8 Remove the 2
2a means 2 times a.
We want just a.
To remove times 2, we divide by 2.
We do it on both sides, so they stay equal.
2 divided by 2 is 1.
So just a is left.
6 divided by 2 is 3.
So a = 3!
9 Write the answers
Only one case works.
So a = 3.
10 Check
Let's check: put a = 3 in.
The left bars give 4.
The right bars give 4 too.
The same!
Let's look back.
First: two cases.
Then: solve the first case.
Then: get a on one side.
Then: solve the second case.
Then: open the brackets.
Then: get a on one side.
Then: remove the + 1.
Then: remove the 2.
Then: write the answers.
Last: check.