Find those x satisfying each of the equations below:

|x + 1| = |x – 5|s

This can be solved in the following cases:

Case1 : when x>-1, |x + 1| = x + 1 and x>5, |x-5| = x-5

x + 1 = x-5 ⇒ no solution as x gets cancelled out on both sides……….eq(1)

Case2 : when x>-1, |x + 1| = x + 1 and x<5, |x-5| = -(x-5)

x + 1 = -(x-5)

⇒ x + 1 = 5-x

⇒ 2x = 5-1

⇒ 2x = 4

⇒ x = 2………………..eq(2)

Case3 : when x<-1, |x + 1| = -(x + 1) and x>5, |x-5| = x-5

-(x + 1) = (x-5)

⇒ -x-1 = x-5

⇒ -2x = -5-1 = -6

⇒ 2x = 6

⇒ x = 3………………..eq(3)

Case4 : when x<-1, |x + 1| = -(x + 1) and x<5, |x-5| = -(x-5)

-(x + 1) = -(x-5)

⇒ -x-1 = -x + 5 ⇒ no solution as x gets cancelled out on both sides…………………..eq(4)

Now from eq(2) ans eq(3), we have x = 2 and x = 3 as the solution of the equation.