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| x + 1 | <= | x - 3 |

With two absolute values, there are 4 possible solutions:

First: x + 1 <= x - 3

Second: x + 1 <= -(x - 3)

Third: -(x + 1) <= x - 3

Fourth: -(x + 1) <= -(x - 3)

Let's work them out one at a time:

.

First: x + 1 <= x - 3

subtract x from both sides

1 <= -3

Never true.  This one can never happen.

.

Second: x + 1 <= -(x - 3)

x +1 <= -x + 3

add x to both sides

2x + 1 <= 3

subtract 1 from both sides

2x < = 2

Answer:  x < = 1

.

Third: -(x + 1) <= x - 3

-x - 1 <= x - 3

add x to both sides

-1 <= 2x - 3

add 3 to both sides

2 <= 2x

divide by 2 on both sides

1 <= x

Answer:  x >= 1

.

Fourth: -(x + 1) <= -(x - 3)

-x - 1 <= -x + 3

add x to both sides

-1 <= 3

Never true.  This solution can't happen.

.

So our 4 solutions are:

Can't happen, x <= 1, x >= 1, can't happen.

Either of the two solutions (the ones that can happen) can be true and the original equation is true.

But our two solutions (x <= 1, x >=1) cover everything- there is no value for x that doesn't fit into one of those solutions.  That means x can be anything.

Answer:  x can be any number.
by Level 13 User (103k points)

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