|2x+1|+1≤5
|2x+1| ≤ 4
-4 ≤ 2x+1 ≤ 4
-5 ≤ 2x ≤ 3
-5/2 ≤ x ≤ 3/2
| 2x + 1 | + 1 ⤠5
2x + 1 + 1 ⤠5
2x ⤠5 - 1 - 1
2x ⤠5 - 2
2x ⤠3
x ⤠3 / 2
Therefore, " x " is equal to 3/2 or less than 3/2
Checking :
| 2 ( 3/2 ) + 1 | + 1 ⤠5
| 6/2 + 1 | + 1 ⤠5
| 3 + 1 | + 1 ⤠5
3 + 1 + 1 ⤠5
5 ⤠5
| 2 ( 2/2 ) + 1 | + 1 ⤠5
| 4/2 + 1 | + 1 ⤠5
| 2 + 1 | + 1 ⤠5
2 + 1 + 1 ⤠5
4 ⤠5 ( The 2/2 is not a solution to the problem )
My Answer :
x = 3/2
|2x+1|+1 ⤠5
subtract 1
|2x+1| ⤠4
this means
2x+1 ⤠4
and
-(2x+1) â¤4
solving the first
2xâ¤3
x ⤠3/2
the second..
-2x-1 ⤠4
-2xâ¤5
x ⥠-5/2
so x ⤠3/2 or x ⥠-5/2
-5/2 ⤠x ⤠3/2
First answer posted is correct:
|2x+1|+1â¤5
2x+1â¤4 or 2x+1â¥-4
2xâ¤3 or 2xâ¥-5
xâ¤3/2 or xâ¥-5/2
The solution is the interval of (-2.5, 1.5)
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Answers & Comments
Verified answer
|2x+1|+1≤5
|2x+1| ≤ 4
-4 ≤ 2x+1 ≤ 4
-5 ≤ 2x ≤ 3
-5/2 ≤ x ≤ 3/2
| 2x + 1 | + 1 ⤠5
2x + 1 + 1 ⤠5
2x ⤠5 - 1 - 1
2x ⤠5 - 2
2x ⤠3
x ⤠3 / 2
Therefore, " x " is equal to 3/2 or less than 3/2
Checking :
| 2x + 1 | + 1 ⤠5
| 2 ( 3/2 ) + 1 | + 1 ⤠5
| 6/2 + 1 | + 1 ⤠5
| 3 + 1 | + 1 ⤠5
3 + 1 + 1 ⤠5
5 ⤠5
Checking :
| 2x + 1 | + 1 ⤠5
| 2 ( 2/2 ) + 1 | + 1 ⤠5
| 4/2 + 1 | + 1 ⤠5
| 2 + 1 | + 1 ⤠5
2 + 1 + 1 ⤠5
4 ⤠5 ( The 2/2 is not a solution to the problem )
My Answer :
x = 3/2
|2x+1|+1 ⤠5
subtract 1
|2x+1| ⤠4
this means
2x+1 ⤠4
and
-(2x+1) â¤4
solving the first
2xâ¤3
x ⤠3/2
the second..
-2x-1 ⤠4
-2xâ¤5
x ⥠-5/2
so x ⤠3/2 or x ⥠-5/2
-5/2 ⤠x ⤠3/2
First answer posted is correct:
|2x+1|+1â¤5
|2x+1| ⤠4
2x+1â¤4 or 2x+1â¥-4
2xâ¤3 or 2xâ¥-5
xâ¤3/2 or xâ¥-5/2
-5/2 ⤠x ⤠3/2
The solution is the interval of (-2.5, 1.5)