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Solve each of the following system of equations in R.
The equation can be re–written as
Adding 1 both the sides, we get,
Subtracting 3 both the sides
Clearly, x≠0, as it will lead equation unmeaningful.
Now, two case arise:
Case1: x+2>0
⇒ x>–2
In this case |x+2|=x+2
Considering Numerator,
2x–2>0
⇒ x>1
⇒ x ϵ (1, ∞) ….(1)
Case 2: x+2<0
⇒ x<–2
In this case, |x+2|=–(x+2)
Considering Numerator,
4x+2>0
ut x<–2
Now, from Denominator, we have–
⇒ x ϵ (–∞ , 0) …(2)
(1, ∞) (from 1 and 2)
We can verify the answers using graph as well.