Explanation:
1st solution
When x-1 is positive
x-1+x≤5
2x-1≤5
2x≤6
x≤3
x-1≥0
x≥1
Solution: x€[1,3]
2nd solution
When x-1 is negative
-(x-1)+x≤5
-x-1+x≤5
-1≤5
0≤5
x-1<0
x<1
Solution: x€(-infinity, 1)
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