Answer:
The answer is: { all those real numbers x | 1<x<4}
Step-by-step explanation:
We have to find the common region for the inequalities:
x+1<5 and x-4>-3 ; where x is a real number.
i.e we need to find the region of ( x + 1 < 5) ∩ ( x - 4 > -3).
let us find the region for : x+1<5
⇒ x<5-1 (subtracting both side by 1)
⇒ x<4
the region is (-∞,4)
in set-builder definition form it could be written as: { all those real number x | -∞<x<4}
now calculating the region for the second inequality: x-4>-3
⇒ x>-3+4 (Adding 4 on both the sides of the inequality)
⇒ x>1
Hence, the region is (1,∞)
in set-builder definition form it could be written as: { all those real number x | 1<x<∞}.
So, the common region in (-∞,4) and (1,∞) i.e. (-∞,4)∩(1,∞)=(1,4).
Hence the answer is: { all those real numbers x | 1<x<4}.
