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Solve each of the following system of equations in R.
Subtracting 3 from both sides, we get–
|x+1|+|x|–3>0
For this, we have 3 cases,
Case 1:–∞ <x<–1
For this, |x+1|=–(x+1) and |x|=–x
–(x+1)–x–3>0
–2x–1–3>0
2x+4<0
x<–2
⇒ x ϵ (–∞ , –2) …(1)
Case 2: –1<x<0
For this, |x+1|=x+1 and |x|=–x
x+1–x–3>0
–2>0
Which is absurd, thus it leads to no solution.
Case 3: 0<x<∞
For this, |x+1|=x+1 and |x|=x
x+1+x–3>0
2x–2>0
x>1
⇒ x ϵ (1 , ∞ ) …(2)
⇒ xϵ (–∞ , –2)⋃ (1 , ∞ ) (from 1 and 2)
We can verify the answers using graph as well.