# Find the integers x if |x-2|=<3.

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### 2 Answers

l x-2 l =< 3

To solve the absolute value, we need to solve fro two cases:

case(1);

x - 2 =< 3

Now we will add 2 to both sides:

==> x =< 5

Then the answer is:

x = { -inf, ....,-1,0,1,2,3,4,5]

Case(2):

-(x-2) =< 3

==> -x + 2 =< 3

Now we will subtract 2 from both sides:

==> -x =< 1

Now we will multiply by -1 and reverse the inequality.

==> x >= -1

==> x = {-1,0,1,2,3.....,inf}

Then, from (1) and (2) we conclude that the answer is:

x = { -inf, ..., 1,2,3,4,5} AND { -1,0,1,2...., inf}

= { -1,0,1,2,3,4,5}

**==> x= { -1,0,1,2,3,4,5}**

From enunciation, we'll have to find out the integer elements that satisfy the inequality |x-2|=<3.

We'll re-write the constraint |x-2|=<3:

-3=< x-2 =<3

We'll solve the left side of the inequality:

-3 =< x-2

We'll add 3 both sides:

0 =<x - 2 + 3

0 =< x + 1

We'll subtract 1 both sides:

-1 =< x

Now, we'll solve the right side:

x - 2 =< 3

We'll add 2 both sides, to isolate x:

x =< 5

So, from both inequalities, we'll get: -1 =< x =< 5

**The integer elements of the set are:****{-1 ; 0 ; 1 ; 2 ; 3 ; 4 ; 5}.**