Q. 1 A3.7( 3 Votes )

# <span lang="EN-US

.|x – 1| = |x – 3|

This can be solved in the following cases:

Case1 : when x>1, |x -1| = x -1 and x>3, |x-3| = x-3

x-1 = x-3 no solution as x gets cancelled out on both sides……….eq(1)

Case2 : when x>1, |x-1| = x -1 and x<3, |x-3| = -(x-3)

x-1 = -(x-3)

x-1 = 3-x

2x = 3 + 1

2x = 4

x = 2………………..eq(2)

Case3 : when x<1, |x-1| = -(x-1) and x>3, |x-3| = x-3

-(x-1) = (x-3)

-x + 1 = x-3

-2x = -3-1 = -4

2x = 4

x = 2………………..eq(3)

Case4 : when x<1, |x-1| = -(x-1) and x<3, |x-3| = -(x-3)

-(x-1) = -(x-3)

-x + 1 = -x + 3 no solution as x gets cancelled out on both sides…………………..eq(4)

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

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