All real numbers x that satisfy ∣2x+1∣<5−∣2x∣ are....
Explanation
Define each absolute value
∣2x+1∣={2x+1,−(2x+1),for 2x+1≥0 or x≥−21for 2x+1<0 or x<−21
∣2x∣={2x,−2x,for 2x≥0 or x≥0for 2x<0 or x<0
It is seen that the boundary values of x from both absolute values are x=−21 and x=0. This means the boundaries will form three regions, namely Region I (x<−21), Region II (−21≤x<0), and Region III (x≥0). Based on the regions and their definitions, the problem can be solved.
Region I: (x<−21)∣2x+1∣=−(2x+1)
∣2x∣=−2x
∣2x+1∣<5−∣2x∣
−(2x+1)<5−(−2x)
−2x−1<5+2x
−4x<6
x>−23
The solution for Region I is {x<−21}∩{x>−23}={−23<x<−21}.
Region II: (−21≤x<0)∣2x+1∣=2x+1
∣2x∣=−2x
∣2x+1∣<5−∣2x∣
2x+1<5−(−2x)
2x+1<5+2x
1<5
All x in Region II are true, so the solution is {−21≤x<0}.
Region III: (x≥0)∣2x+1∣=2x+1
∣2x∣=2x
∣2x+1∣<5−∣2x∣
2x+1<5−2x
4x<4
x<1
The solution is {x≥0}∩{x<1}={0≤x<1}.
So the overall solution is the union
{−23<x<−21}∪{−21≤x<0}∪{0≤x<1}={−23<x<1}