Solve ( x + 1 < 5) ∩ ( x - 4 > -3). {all real numbers} { x | 1 < x < 4} { x | x < 4 or x > 1}

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}.

because they could drop even lower..